At One Time The Thickness Of Different Gauges Of Stainless Steel Were Given In Fractions Of An Inch. (2024)

The standard deviation of a sample portion is

[tex]\sigma_{\hat{p}}=\sqrt[]{\frac{pq}{n}}[/tex]

Where p = 45 and q = 1 - p.

[tex]q=1-0.45=0.55[/tex]

Let's find the standard deviation of the sample n = 75.

[tex]\sigma_{\hat{p}}=\sqrt[]{\frac{0.45\cdot0.55}{75}}\approx0.057[/tex]The standard deviation of the first sample is 0.057.

Repeat the process for n = 1500.

[tex]\sigma_{\hat{p}}=\sqrt[]{\frac{0.45\cdot0.55}{1500}}\approx0.0128[/tex]The standard deviation of the second sample is 0.0128.

Therefore,

• When n = 75, the standard deviation is 0.057.

,

• When n = 1500, the standard deviation is 0.0128.

Remember that

we have 3 prime numbers (2,3,5)

and

we have two composite numbers on a 6 sided die (4,6)

therefore

The probability of getting a prime number is P=3/6=1/2=0.5

The probability of getting a composite number is P=2/6=1/3=0.33

therefore

The more likely outcome is a prime number

g(x) = 35 + 8x

t(x) = 13x + 5

g(x) = t(x)

35 + 8x = 13x + 5 (Replacing)

35 - 5 + 8x = 13x (Subtracting 5 from both sides of the equation)

35 - 5= 13x - 8x (Subtracting 8x from both sides of the equation)

30= 5x (Subtracting)

30/5 = x (Dividing by 5 on both sides of the equation)

6= x

The answer is x= 6

The question is given as :

[tex]3x^2-2=25[/tex]

collect like terms as;

[tex]3x^2=25+2[/tex][tex]3x^2=27[/tex]

Divide both sides by 3 as

[tex]x^2=\frac{27}{3}[/tex][tex]\begin{gathered} x=\sqrt[\square]{9} \\ \end{gathered}[/tex][tex]x=3[/tex]

Solution:

Given the expression below

[tex]4|8x-2|=8[/tex]

Applying the absolute rule, which is

[tex]if\text{ \mid u\mid=a, a}>0,\text{ then u}=-a\text{ or u}=a[/tex]

Applying the rule above, the expression becomes

[tex]\begin{gathered} 4|8x-2|=8 \\ Divide\text{ both sides by 4} \\ |8x-2|=2 \\ -2=8x-2=2 \end{gathered}[/tex]

Hence, a new equivalent equation that is easier to solve is

[tex]-2=8x-2=2[/tex]

We are required to graph out the equations from A and B.

In order to graph out the equations, we need a table of values for our x and y axes.

We can choose the values for our x-axis to be from 0 to 14 moving in steps of 2.

While there will be two corresponding y-values. One for y = 7x and the other for y = 30 + 4x

Therefore, we need to calculate these y values which we shall denote y1 and y2 respectively.

This is done below:

[tex]\begin{gathered} \text{when x = 0;} \\ y_1=7\times0=0 \\ y_2=30+4(0)=30 \\ \\ \text{when x = 2;} \\ y_1=7\times2=14 \\ y_2=30+4(2)=38 \\ \\ when\text{ x = 4;} \\ y_1=7\times4=28 \\ y_2=30+4(4)=46 \\ \\ \text{when x= 6;} \\ y_1=7\times6=42 \\ y_2=30+4(6)=54 \\ \\ \text{when x = 8;} \\ y_1=7\times8=56 \\ y_2=30+4(8)=62 \\ \\ \text{when x= 10;} \\ y_1=7\times10=70 \\ y_2=30+4(10)=70 \\ \\ \text{when x= 12;} \\ y_1=7\times12=84 \\ y_2=30+4(12)=78 \\ \\ \text{when x = 14;} \\ y_1=7\times14=98 \\ y_2=30+4(14)=86 \end{gathered}[/tex]

The calculated coordinates are:

(x, y1), (x, y2):

(0, 0), (0, 30)

(2, 14)(2, 38)

(4, 28)(2, 46)

(6, 42)(6, 54)

(8, 56)(8, 62)

(10, 70)(10, 70)

(12, 84)(12, 78)

(14, 98)(14, 86)

Now that we have calculated the values of the coordinates, we can create the table:

Now that we have the table of values, we can now plot the values:

The x values on the x-axis and the y-values on the y-axis

A picture of the plot is shown below:

Notice that on the table of values, both y1 and y2 have the same y-value at x = 10. This is the same point in which the graphs of y1 and y2 intersect i.e. at point x = 10, y = 70

[tex]\begin{gathered} -2n-8>-8 \\ -2n>-8+8 \\ -2n>0 \\ n>\frac{0}{-2} \\ n>0 \end{gathered}[/tex]

Therefore n>0 or

[tex]n\in(0,\infty)[/tex]

To find the mean, we have to use the following formula

[tex]\bar{x}=\frac{\Sigma(x\cdot f)}{N}[/tex]

Where x represents each score and f represents the number of students. Using the given table, we have

[tex]\begin{gathered} \bar{x}=\frac{65\cdot8+70\cdot1+75\cdot2+80\cdot1+85\cdot1+90\cdot9+95\cdot4}{8+1+2+1+1+9+4} \\ \bar{x}=\frac{520+70+150+80+85+810+380}{26}=\frac{2095}{26} \\ \bar{x}\approx80.6 \end{gathered}[/tex]Hence, the mean is 80.6, approximately.

see the attached figure to better understand the problem

Applying the Pythagorean Theorem

l^2=7.7^2+2.6^2

l=8.12712 ft

SOLUTION

(8). From the question, Jenny rode 8km, changed direction and rode another 15km and then turned, by changing direction and rode 17km before checking her map.

Could her path be a right angle?

For her path to be a right angle, then the square of the longest path (17km) must be equal to the sum of the squares of the other two paths (8km and 15km). That is, according to Pythagorean theorem.

Let's check

[tex]\begin{gathered} 17^2=289 \\ 8^2+15^2=64+225=289 \end{gathered}[/tex]

So, since we got 289 as the sum of 8 square and 15 square,

Hence her path is a right-angle. YES

The right-angle is shown below

Following the arrow indicating the direction she moved, we can see that Jenny will be back at her starting point.

Could Jenny be back at her starting point?

The answer is YES

Let's use the following formula:

[tex]v=\frac{d}{t}[/tex]

We have two speeds in this case, the speed of the boat and the speed of the water, so:

[tex]\begin{gathered} v_b=Speed_{\text{ }}of_{\text{ }}the_{\text{ }}bo_{}at \\ v_w=Speed_{\text{ }}of_{\text{ }}the_{\text{ }}water \end{gathered}[/tex]

When the boat travels 33 miles downstream in 4 hours, we can say that their speeds add up:

[tex]v_b+v_w=\frac{33}{4}[/tex]

When the boat return, we can say that their speeds are subtracted:

[tex]v_b-v_w=\frac{33}{4}[/tex]

with this we can form a system of equations 2x2:

[tex]\begin{gathered} v_b+v_w=\frac{33}{4}_{\text{ }}(1)_{} \\ v_b-v_w=\frac{33}{4}_{\text{ }}(2) \end{gathered}[/tex]

Let's solve for vb:

[tex]\begin{gathered} (1)+(2) \\ v_b+v_b+v_w-v_w=\frac{33}{4}+\frac{33}{4} \\ 2v_b=\frac{66}{4} \\ 2v_b=\frac{33}{2} \\ v_b=\frac{33}{4}=\frac{8.25mi}{h} \end{gathered}[/tex]

Answer:

The product is -52

The result is the same with 4.(-13) because multiplication is cummutative.

Explanation:

-13 × 4 = -52

This is the same as

4 × -13

because multiplication is cummutative.

For any numbers a and b,

a × b = b× a

From our table, we have

[tex]P(\text{regular lenses)=}\frac{2+4+4}{19}=\frac{10}{19}[/tex]

similarly

[tex]P(\text{cat eye frame)=}\frac{2+2}{19}=\frac{4}{19}[/tex]

and the searched probability is

[tex]P(\text{regular lenses }\cup\text{ cat eye frame)=}P(\text{regular lenses)}+P(\text{ cat eye frame)-}P(\text{regular lenses }\cap\text{ cat eye frame)}[/tex]

which gives

[tex]\begin{gathered} P(\text{regular lenses }\cup\text{ cat eye frame)=}\frac{10}{19}+\frac{4}{19}-\frac{2}{19} \\ P(\text{regular lenses }\cup\text{ cat eye frame)=}\frac{12}{19} \end{gathered}[/tex]

then, the answer is

[tex]P(\text{regular lenses }\cup\text{ cat eye frame)=}\frac{12}{19}[/tex]

The answer is False. When any data value in a distribution provided is multiplied by a factor, the center and measure of spread of the graph will increase proportionally.

Center and Spread of the graph are a part of summary statistics because they can shortly define the data sets provided. A specific value in a given data set is represented by the Center. The Center of a graph can be Mean, or Median, or Mode.The variation observed in the data set is represented by the Spread. The Spread of a graph can be Range or Standard Deviation.

Now that we have an idea about the Center and Spread of the graph, it can be understood that -

When the values in a data set are multiplied by a factor -

There will be increase in the value of the mean or median and thus, correspondingly, the center of the graph.There will be increase in the value of variation of the data set and thus, correspondingly, the spread of the graph.

Thus, the answer is false. Whenever the data values in a distribution are multiplied by a factor, the center and measure of spread of the graph will NOT increase proportionally.

To learn more about distribution visit https://brainly.com/question/14926605

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First, we find the equation of the function B:

The equation is of the form y = mx + b, where m is:

[tex]m=\frac{y2-y1}{x2-x1}[/tex]

We have the points P1(0,2) and P2(4,0), so

[tex]m=\frac{0-2}{4-0}=-\frac{2}{4}=-\frac{1}{2}[/tex]

And b is:

[tex]\begin{gathered} y=mx+b \\ \text{we use the point (0,2)} \\ 2=-\frac{1}{2}(0)+b \\ b=2 \end{gathered}[/tex]

Therefore, the function B is:

[tex]y=-\frac{1}{2}x+2[/tex]

Next, we solve the system of equations:

[tex]\begin{gathered} y=-3x+2 \\ y=-\frac{1}{2}x+2 \end{gathered}[/tex]

Substitute y:

[tex]-3x+2=-\frac{1}{2}x+2[/tex]

Solve for x:

[tex]\begin{gathered} -3x+\frac{1}{2}x+2=-\frac{1}{2}x+\frac{1}{2}x+2 \\ -\frac{5}{2}x+2=2 \\ -\frac{5}{2}x+2-2=2-2 \\ -\frac{5}{2}x=0 \\ x=0 \end{gathered}[/tex]

Then for y:

[tex]y=-3(0)+2=0+2=2[/tex]

Answer:

x = 0

y = 2

Answer:[tex]V\text{ = }66.7m^3[/tex]Explanations:

The volume of a square based pyramid is given by the formula:

[tex]\begin{gathered} V\text{ = }\frac{a^2h}{3} \\ \text{where V is the volume } \\ a\text{ is the }base\text{ edge} \\ \text{and h is the height} \end{gathered}[/tex][tex]\begin{gathered} V\text{ = }\frac{5^2\times8}{3} \\ V\text{ = }\frac{25\times8}{3} \\ V\text{ = }\frac{200}{3} \\ V\text{ = }66.7m^3 \end{gathered}[/tex]

Given data:

The expression for the population of herds is y=18.21(1.29)^x.

Substitute 65 for y in the above expression.

[tex]\begin{gathered} (65)=18.21(1.29)^x \\ (1.29)^x=3.5695 \\ x\ln (1.29)=\ln (3.5695) \\ x=4.99 \\ x\approx5\text{ years} \end{gathered}[/tex]

Thus, the value of times in year are , so option F is correct.

Hello there. To solve this question, we need to remember some properties on functions and their graphs.

Starting with the function f(x) = x², we want to find a function g(x) such that it represents a shifted to the right and compressed vertically version of f(x).

First, graph f(x) and see what you want to have:

Shift a quadratic function to the right means changing the place of its vertex. In fact, for f(x), we can subtract values inside the square, or complete the square and then do it.

Notice that (x - a)² is a quadratic function with root a. If a is greater than 0, we have the case of the graph below:

So, when we want to shift to the right, you subtract a positive number. If you want to shift it to the left, subtract a negative number, thus it becomes a plus.

Now, talking about compressing it. We'd have something like follows:

This happens when we multiply the function by a factor between 0 and 1.

If the factor is less than 0, it will turn the parabola upside down (works for all types of graphs) and if it is greater than 1, it'll stretch the parabola, becoming bigger and not compressed.

In the options we have, the only option that satisfies both compressed vertically and shifted to the right version of f(x) is G(x) = 1/2 * (x - 6)².

Given

The equation,

[tex]8y^2=12y[/tex]

To solve for y.

Explanation:

It is given that,

[tex]8y^2=12y[/tex]

That implies,

[tex]\begin{gathered} 8y^2-12y=0 \\ 4y(2y-3)=0 \\ 4y=0,\text{ }2y-3=0 \\ y=0,\text{ }y=\frac{3}{2} \end{gathered}[/tex]

Hence the value of y is y = 0,3/2.

Solution: Better buy $ 19.24 for 20 packs of gum.

Analysis: As we want to compare two purchase options of packs of gum, we need to find the rate of each option. In this case, we want to find the rate: Price per 1 pack of gum. We find it calculating the total price between the number of packs of gum.

Option 1:

[tex]Rate=\frac{29.38}{30}=0.9793\cong0.98[/tex]

Option 2:

[tex]Rate:\text{ }\frac{19.24}{20}=0.962\cong0.96[/tex]

After we find both rates, let's compare them:

[tex]0.98>0.96[/tex]

For that reason, we can assume the best purchase option is $ 19.24 for 20 packs of gum.

D. I and II

The values that satisfies this condition

[tex]f(x)\ge g(x)[/tex]

Are the values of y, or the Range that are higher than the values of g(x).

For these functions. we have

I. -2 ≤ x ≤ 0 True

Since according the graph all the values within this interval, f(x) is greater than or equal than f(x).

Because when x=-1 y= 4 for f(x)

for g(x) x=-1 y=2

II. 0 ≤ x ≤ 1 true

Because

f(x) when x= 0 y=6

g(x) when x=0 y=4

III. 1 ≤ x ≤ 3

f(x) when x= 1 y=6

g(x) when x=1 y=6

Given the expression:

[tex]-\frac{3}{4}x-4[/tex]

We evaluate this expression for x = 0:

[tex]\begin{gathered} -\frac{3}{4}x-4=-\frac{3}{4}(0)-4=0-4 \\ \\ \therefore-4 \end{gathered}[/tex]

Answer: -4

SOLUTION

Write out the given information.

[tex]\begin{gathered} \text{radius,r}=10\operatorname{cm} \\ \theta=4\pi\text{rad} \end{gathered}[/tex]

The length of arc with radius and angle in radians is given by

[tex]\begin{gathered} length\text{ of an arc=}\theta\times r \\ \text{Where }\theta\text{ is the angle in rads} \end{gathered}[/tex]

Substituting this value into the formula, we have:

[tex]\text{Length of an arc=4}\pi\times10=40\pi[/tex]

Therefore, the length of arc is 40π rad

Answer:40π rad.

Input data

2 type of coins

Customer 1

8 coins, 3 of which are nickles

Customer 2

11 coins, 7 of which are nickles

Procedure

V = total value

Customer 1

V = (8-3)c + (3)*0.05

Customer 2

V = (11-7)c + (7)*0.05

Now we are going to equal the equations

[tex]\begin{gathered} 5c+3\cdot0.05=4c+7\cdot0.05 \\ 5c-4c=7\cdot0.05-3\cdot0.05 \\ c=4\cdot0.05 \\ c=\text{0}.20 \end{gathered}[/tex]

The coin c would be c = 0.20

Solution:

The continuous compound interest is expressed as

[tex]\begin{gathered} P(t)=P_0\times e^{rt} \\ where \\ P(t)=value\text{ at time t} \\ P_0=principal\text{ amount} \\ r=annual\text{ interest rate} \\ t=length\text{ of time the interest is applied} \end{gathered}[/tex]

Given that

[tex]\begin{gathered} t=6 \\ r=3.5\%=0.035 \\ P_0=\$16000 \end{gathered}[/tex]

By substitution, we have

[tex]\begin{gathered} P(t)=16000\times e^{(0.035\times6)} \\ =19738.84895 \\ \Rightarrow P(t)\approx\$19738.8\text{ \lparen nearest cent\rparen} \end{gathered}[/tex]

Hence, we have

Explanation

Question 5

(2,-8)

Explanation

The rule for a rotation by 180° about the origin is:

[tex](x,y)\Rightarrow(-x,-y)[/tex]

in other words you have to change the sign of x component and y component, so

[tex](-2,8)\Rightarrow rotated\text{ 180 \degree}\Rightarrow(-(-2),-(8))=(2,-8)[/tex]

so, the point that results is

(2,-8)

I hope this helps you

1 < 1.2

This may be interpreted as 1 is less than 1.2

According to the problem, the event is "the outcome is greater than 2", so the probability is

[tex]P(A)=0.3+0.3=0.6[/tex]Hence, event A has 0.6 chances.

Answer:

Explanation:

From the graph, we can read off the vertices of triangle KLM as;

[tex]\begin{gathered} K(0,4) \\ L(9,4) \\ M(-1,-6) \end{gathered}[/tex]

To translate 1 unit right, we have to add 1 to each of the x-coordinates of the vertices of triangle KLM.

To translate 4 units down, we have to subtract 4 from each of the y-coordinates of the vertices of triangle KLM.

So we'll have;

[tex]K^{\prime}(0+1,4-4)\rightarrow K^{\prime}(1,0)[/tex][tex]L^{\prime}(9+1,4-4)\rightarrow L^{\prime}(10,0)[/tex][tex]M^{\prime}(-1+1,-6-4)\rightarrow M^{\prime}(0,-10)[/tex]

Therefore, the graph of t