e
65. the perpendicular
bisector of the
segment with
endpoints (-5/2,-2)
and (3, 5)
HELP PLEASE! Picture included!

Answer:

The volume of the crystal is [tex]V=125 \:cm^3[/tex].

Step-by-step explanation:

The volume enclosed by a cube is the number of cubic units that will exactly fill a cube.

To find the volume of a cube recall that a cube has all edges the same length. The volume of a cube is found by multiplying the length of any edge by itself three times. Or as a formula

                                                         [tex]V=s^3[/tex]

where:

s is the length of any edge of the cube.

From the information given we know that the crystal has edge lengths of 5 centimeters. Therefore, the volume of the crystal is

[tex]V=5^3=125 \:cm^3[/tex].

Answer:

[tex]the \: answer \: is \: d.(x + 4) {}^{2} = 8(y + 4)[/tex]

Answer:

Graph 2

Step-by-step explanation:

You can see that all the shaded numbers are above negative 25 in that graph. Hope this helped!

Answer:

  x = 8

Step-by-step explanation:

The angle with the expression in it is complementary to the 30° angle, so is 60°. Then we have ...

  4 +7x = 60

  7x = 56 . . . . . . subtact 4

  x = 8 . . . . . . . . .divide by 7

Answer 9 is in the domain of f g

Step-by-step explanation:

Answer:

Step-by-step explanation:

Hello!

An ANOVA was conducted to analyze the variable

Y: sales of "People" magazine over a 5-week period

This was studied in 4 Borders outlets in Chicago.

So this test has one factor: "Borders outlets" and four treatments: "Store 1, store 2, store 3 and store 4"

For each store you have the data for the weekly sales over a 5-week period so the sample sizes are:

n₁=n₂=n₃=n₄= 5 weeks

Store 1

∑X₁= 540; ∑X₁²= 58434

X[bar]₁= 108

S₁²= 28.50

S₁= 5.34

Store 2

∑X₂= 437; ∑X₂²= 38663

X[bar]₂= 87.40

S₂²= 117.30

S₂= 10.83

Store 3

∑X₃= 455; ∑X₃²= 41899

X[bar]₃= 91

S₃²= 123.50

S₃= 11.11

Store 4

∑X₄=505; ∑X₄²= 51429

X[bar]₄= 101

S₄²= 106

S₄= 10.30

Totals

N= n₁ + n₂ + n₃ + n₄= 4*5= 20

∑Mean= 108+87.40+91+101= 387.4

∑Variance= 28.50+117.30+123.50+106= 375.30

∑Standard deviation= 5.34+10.83+11.11+10.30= 37.58

Hypothesis test:

H₀: μ₁= μ₂= μ₃= μ₄

H₁: At least one population mean is different.

α: 0.05

The statistic for this test is

[tex]F= \frac{MS_{Treatments}}{MS_{Error}} ~~F_{k-1;N-k}[/tex]

k-1= 3 Df of the treatments, k=4 number of treatments

N-k= 16 Df of errors, N=20 total number of observations in all treatments

[tex]F_{H_0}= \frac{441.78}{93.83}= 4.71[/tex]

The critical region and p-value for this test are one-tailed to the right.

Using the critical value approach:

[tex]F_{k-1;N-k;1-\alpha }= F_{3;16;0.95}= 3.25[/tex]

The decision rule is:

If [tex]F_{H_0}[/tex] ≥ 3.25, reject the null hypothesis.

If [tex]F_{H_0}[/tex] < 3.25, do not reject the null hypothesis.

Using the p-value approach:

Little reminder: The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).

So under the distribution F₃,₁₆ you have to calculate the probability of the calculated  [tex]F_{H_0}[/tex]:

P(F₃,₁₆≥4.71)= 1 - P(F₃,₁₆<4.71)= 1 - 0.9847 = 0.0153

p-value= 0.0153

The decision rule for this approach is

If p-value ≤ α, reject the null hypothesis.

If p-value > α, do not reject the null hypothesis.

The p-value is less than the level of significance, so the decision is to reject the null hypothesis.

I hope this helps!

Answer:

[tex] (x+11)^2 = (x+3)^2 +16^2[/tex]

And if we solve this equation for x we got:

[tex] x^2 +22x +121 = x^2 +6x +9 +256[/tex]

We can cancel [tex]x^2[/tex] in both sides and we have this:

[tex] 22x -6x= 256+9-121 =144[/tex]

And then we got:

[tex] 16 x= 144[/tex]

[tex] x =\frac{144}{16}= 9[/tex]

And then the length of the sides are 9+11= 20 m for the hypothenuse, 16 for the adjacent side and 9+3 = 12m for the last side.

Lenght of the smaller unknown side: 12m

Lenght of the larger unknown side: 20m

Step-by-step explanation:

For this case we have a right triangle and we can use the Pythagoras Theorem and using the info given by the triangle we can set up the following equation:

[tex] (x+11)^2 = (x+3)^2 +16^2[/tex]

And if we solve this equation for x we got:

[tex] x^2 +22x +121 = x^2 +6x +9 +256[/tex]

We can cancel [tex]x^2[/tex] in both sides and we have this:

[tex] 22x -6x= 256+9-121 =144[/tex]

And then we got:

[tex] 16 x= 144[/tex]

[tex] x =\frac{144}{16}= 9[/tex]

And then the length of the sides are 9+11= 20 m for the hypothenuse, 16 for the adjacent side and 9+3 = 12m for the last side side.

Lenght of the smaller unknown side: 12m

Lenght of the larger unknown side: 20m

Answer:

[tex] Median = \frac{60+60}{2}=60[/tex]

And we see that the closest values to 60 are 62 and 63 and then the answer would be:

The values closest to the middle are 62 inches and 63 inches.

Step-by-step explanation:

We have the following dataser given:

62 33 50 90  80 50 40 70  50 63 74 53  55 64 60 60  78 70 43 82

We can sort the values from the lowest to the highest and we got::

33 40 43 50 50 50 53 55 60 60 62 63 64 70 70 74 78 80 82 90

Now we see that we have n=20 values and the values closest to the middle and we can use the middle as the median and for this case the median can be calculated from position 10 and 11th and we got:

[tex] Median = \frac{60+60}{2}=60[/tex]

And we see that the closest values to 60 are 62 and 63 and then the answer would be:

The values closest to the middle are 62 inches and 63 inches.

The values closest to these middle elements are 60 and 63 inches

The dataset is given as:

62 33 50 90 80 50 40 70 50 63 74 53 55 64 60 60 78 70 43 82

Next, we sort the data elements in ascending order

33 40 43 50 50 50 53 55 60 60 62 63 64 70 70 74 78  80 82 90

The length of the dataset is 20.

So, the elements at the middle are the 10th and the 11 elements.

From the sorted dataset, these elements are: 60 and 62

Hence, the values closest to these middle elements are 60 and 63

Read more about median at:

https://brainly.com/question/14532771

Answer:

A

Step-by-step explanation:

They are congruent because of the SSS theorem.  The chords are congruent because the angles are congruent (angles are congruent because they are vertical angles).  Congruent central angles have congruent chords.  The other two sides are congruent because they are all radii of the circle and radius are always congruent.

Answer:

A

Step-by-step explanation:

The triangles are isosceles and congruent  too. ( both have 2  sides congruent  as  radius and the angles between them are congruent- SAS)

Answer:

-7,9

Step-by-step explanation:

x-1=-8

x=-7

x-1=8

x=9

Answer:

f(x)=√x

Step-by-step explanation:

A parent function is something simple with just x

Answer:

f(x)=Square Root of x (choice B or the second one down)

Step-by-step explanation:

All the other choices have either a plus 3 or minus 3. A parent function is not going to have any type of number being added or subtracted to it.

The slope is zero. Slope formula is Y=mx+b and since B is 1.5 and it is a straight line, Y=mx+1.5. What plus 1.5 is 1.5? 0. Hope this helps.

Hope this helps!

Stay safe, have a good day :D

Answer:

a) [tex]x(t) = 13*e^(^-^\frac{t}{100}^)[/tex]

b) 10.643 kg

Step-by-step explanation:

Solution:-

- We will first denote the amount of salt in the solution as x ( t ) at any time t.

- We are given that the Pure water enters the tank ( contains zero salt ).

- The volumetric rate of flow in and out of tank is V(flow) = 50 L / min  

- The rate of change of salt in the tank at time ( t ) can be expressed as a ODE considering the ( inflow ) and ( outflow ) of salt from the tank.

- The ODE is mathematically expressed as:

                            [tex]\frac{dx}{dt} =[/tex] ( salt flow in ) - ( salt flow out )

- Since the fresh water ( with zero salt ) flows in then ( salt flow in ) = 0

- The concentration of salt within the tank changes with time ( t ). The amount of salt in the tank at time ( t ) is denoted by x ( t ).

- The volume of water in the tank remains constant ( steady state conditions ). I.e 10 L volume leaves and 10 L is added at every second; hence, the total volume of solution in tank remains 5,000 L.

- So any time ( t ) the concentration of salt in the 5,000 L is:

                             [tex]conc = \frac{x(t)}{1000}\frac{kg}{L}[/tex]

- The amount of salt leaving the tank per unit time can be determined from:

                         salt flow-out = conc * V( flow-out )  

                         salt flow-out = [tex]\frac{x(t)}{5000}\frac{kg}{L}*\frac{50 L}{min}\\[/tex]

                         salt flow-out = [tex]\frac{x(t)}{100}\frac{kg}{min}[/tex]

- The ODE becomes:

                               [tex]\frac{dx}{dt} = 0 - \frac{x}{100}[/tex]

- Separate the variables and integrate both sides:

                       [tex]\int {\frac{1}{x} } \, dx = -\int\limits^t_0 {\frac{1}{100} } \, dt + c\\\\Ln( x ) = -\frac{t}{100} + c\\\\x = C*e^(^-^\frac{t}{100}^)[/tex]

- We were given the initial conditions for the amount of salt in tank at time t = 0 as x ( 0 ) = 13 kg. Use the initial conditions to evaluate the constant of integration:

                              [tex]13 = C*e^0 = C[/tex]

- The solution to the ODE becomes:

                           [tex]x(t) = 13*e^(^-^\frac{t}{100}^)[/tex]

- We will use the derived solution of the ODE to determine the amount amount of salt in the tank after t = 20 mins:

                           [tex]x(20) = 13*e^(^-^\frac{20}{100}^)\\\\x(20) = 13*e^(^-^\frac{1}{5}^)\\\\x(20) = 10.643 kg[/tex]

- The amount of salt left in the tank after t = 20 mins is x = 10.643 kg

Use the Pythagorean theorem to solve.

Height = sqrt(12^2 -3^2)

Height = sqrt(144-9)

Height = sqrt(135)

Height = 11.6189 = 11.6 feet

Answer:

[tex]E_{A} =0.25*160=40[/tex]  

[tex]E_{B} =0.25*160=40[/tex]

[tex]E_{C} =0.4*160=64[/tex]

[tex]E_{D} =0.05*160=8[/tex]  

[tex]E_{F} =0.05*160=8[/tex]  

And now we can calculate the statistic:  

[tex]\chi^2 = \frac{(36-40)^2}{40}+\frac{(42-40)^2}{40}+\frac{(60-64)^2}{64}+\frac{(14-8)^2}{8}+\frac{(8-8)^2}{8} =5.25[/tex]  

The answer would be:

c. 5.25

Step-by-step explanation:

The observed values are given by:

A: 36

B: 42

C: 60

D: 14

E: 8

Total =160

We need to conduct a chi square test in order to check the following hypothesis:  

H0: There is no difference in the proportions for the final grades

H1: There is a difference in the proportions for the final grades

The level of significance assumed for this case is [tex]\alpha=0.01[/tex]  

The statistic to check the hypothesis is given by:  

[tex]\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]  

Now we just need to calculate the expected values with the following formula [tex]E_i = \% * total[/tex]  

And the calculations are given by:  

[tex]E_{A} =0.25*160=40[/tex]  

[tex]E_{B} =0.25*160=40[/tex]

[tex]E_{C} =0.4*160=64[/tex]

[tex]E_{D} =0.05*160=8[/tex]  

[tex]E_{F} =0.05*160=8[/tex]  

And now we can calculate the statistic:  

[tex]\chi^2 = \frac{(36-40)^2}{40}+\frac{(42-40)^2}{40}+\frac{(60-64)^2}{64}+\frac{(14-8)^2}{8}+\frac{(8-8)^2}{8} =5.25[/tex]  

The answer would be:

c. 5.25

Now we can calculate the degrees of freedom for the statistic given by:  

[tex]df=(categories-1)=(5-1)=4[/tex]

And we can calculate the p value given by:  

[tex]p_v = P(\chi^2_{4} >5.25)=0.263[/tex]  

The p value is higher than the significance so we have enough evidence to FAIL to reject the null hypothesis

Answer:

The probability that if Sunday is sunny, then Tuesday is also sunny is 0.86.

Step-by-step explanation:

Let us denote the events as follows:

Event 1: a sunny day

Event 2: a rainy day

From the provided data we know that the transition probability matrix is:

                 [tex]\left\begin{array}{ccc}1&\ \ \ \ 2\end{array}\right[/tex]

[tex]\text{P}=\left\begin{array}{c}1&2\end{array}\right[/tex]  [tex]\left[\begin{array}{cc}0.90&0.10\\0.50&0.50\end{array}\right][/tex]

In this case we need to compute that if Sunday is sunny, what is the probability that Tuesday is also sunny.

This implies that we need to compute the value of P₁₁².

Compute the value of P² as follows:

[tex]P^{2}=P\cdot P[/tex]

     [tex]=\left[\begin{array}{cc}0.90&0.10\\0.50&0.50\end{array}\right]\cdot \left[\begin{array}{cc}0.90&0.10\\0.50&0.50\end{array}\right]\\\\=\left[\begin{array}{cc}0.86&0.14\\0.70&0.30\end{array}\right][/tex]

The value of P₁₁² is 0.86.

Thus, the probability that if Sunday is sunny, then Tuesday is also sunny is 0.86.

Answer:

Step-by-step explanation:

[tex]y = f(x) =\frac{1}{\sqrt{3 \pi} } e^{-x^{2/3}}[/tex]

y = 0, x = 0 and x = 3

Consider an element of thickness dx at a distance x from the origin. By Cylindirical Shell Method, the volume of the element is given by

[tex]dV=(2\pi rdr)h=(2\pi xdx)f(x) => dV=(2\pi xdx) \frac{1}{\sqrt{3\pi}}e^{-x^{\frac{2}{3}}}[/tex]

[tex]dV=2\sqrt{\frac{\pi}{3}}xe^{-x^{\frac{2}{3}}}dx[/tex]

Integrate the above integral over the limits x=0 to x=3 which implies

[tex]\int_{0}^{V}dV=2\sqrt{\frac{\pi}{3}}\int_{0}^{3}xe^{-x^{\frac{2}{3}}}dx[/tex]

Solve by subsititution

[tex]Let,\\ -x^{\frac{2}{3}}=y => \frac{-2}{3}x^{\frac{-1}{3}}dx=dy => x^{\frac{-1}{3}}dx=\frac{-3}{2}dy[/tex]

Also, apply the new limits

[tex]At,\\\\ x=0, y=0 \ and \ At, x=3, y=-\sqrt[3]{9}[/tex]

This implies,

[tex]\int_{0}^{V}dV=2\sqrt{\frac{\pi}{3}}\int_{0}^{3}x^{\frac{4}{3}}e^{-x^{\frac{2}{3}}}x^{\frac{-1}{3}}dx=2\sqrt{\frac{\pi}{3}}\int_{0}^{-\sqrt[3]{9}}y^{2}e^{y}(\frac{-3}{2})dy[/tex]

[tex]V=-\sqrt{3\pi}\int_{0}^{-\sqrt[3]{9}}y^{2}e^{y}dy[/tex]

Let,

[tex]I=\int_{0}^{-\sqrt[3]{9}}y^{2}e^{y}dy[/tex]

Integrate by parts the above integral

[tex]u=y^2 \ and \ dv=e^ydy => du=2y \ and \ v=e^y[/tex]

Integrate by parts formula

[tex]\int udv=uv-\int vdu => y^2e^y-\int 2ye^ydy[/tex]

Again integrate by parts

[tex]u=y \ and \ dv=e^ydy => du=1 \ and \ v=e^y[/tex]

Integrate by parts formula

[tex]\int udv=uv-\int vdu => y^2e^y-2[ye^y-e^y]=e^y[y^2-2y+2][/tex]

Therefore,

[tex]I=[e^y(y^2-2y+2)]_{0}^{-\sqrt[3]{9}}\\\\=e^{-2.0802}[(2.0802)^2+2(2.0802)+2]-e^{0}[0-0+2]\\\\\frac{(4.3272+4.1604+2)}{8.0061}-2\\\\=\frac{10.4876}{8.0061}-2\\\\=1.3099-2\\\\=-0.6901[/tex]

This implies, the volume is

[tex]V=-\sqrt{3\pi}I\\\\=-\sqrt{3\times 3.142} \times (-0.6901)\\\\=3.0701 \times 0.6901\\\\=2.1186[/tex]

That is, up to three decimal places

[tex]V\approx 2.118[/tex]

Answer:

c) The Z-score = - 1.90 unusual

Step-by-step explanation:

Explanation:-

Let 'X' be the random variable in normal distribution

Given student received a score X = 50

Mean of the Population x⁻ = 69

standard deviation of the Population 'σ' = 10

now

[tex]Z = \frac{x^{-}-mean }{S.D}[/tex]

[tex]Z = \frac{x^{-}-mean }{S.D} = \frac{50 -69}{10} = - 1.90[/tex]

The Z-score = - 1.90

Conclusion:-

The Z-score = - 1.90 unusual

Answer: The Z-score = - 1.90 unusual

Step-by-step explanation:

Answer:

n = -1

Step-by-step explanation:

So first subtract 9 to both sides

8n = -n - 9

Now you want the n on one side and the constant on the other

so add the single n to the n side

9n = -9

Divide 9 to both sides to solve for n

n = -1

Answer:

61.8

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

[tex]\mu = 172.6, \sigma = 42.4[/tex]

What percentage of women have weights between the ejection seat’s weight limits (that is, 133.8 to 208.0 lb)?

We have to find the pvalue of Z when X = 208 subtracted by the pvalue of Z when X = 133.8 for the proportion. Then we multiply by 100 to find the percentage.

X = 208

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{208 - 172.6}{42.4}[/tex]

[tex]Z = 0.835[/tex]

[tex]Z = 0.835[/tex] has a pvalue of 0.798

X = 133.8

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{133.8 - 172.6}{42.4}[/tex]

[tex]Z = -0.915[/tex]

[tex]Z = -0.915[/tex] has a pvalue of 0.180

0.798 - 0.18 = 0.618

0.618*100 = 61.8%

Without the %, the answer is 61.8.

Answer:

Option (2). 14 m

Step-by-step explanation:

Formula to get the area of a circle 'A' = [tex]\pi r^{2}[/tex]

where r = radius of the circle

Given in the question,

Area of the circle = 153.86 square meters

By putting the values in the formula,

153.86 = πr²

r = [tex]\sqrt{\frac{153.86}{\pi } }[/tex]

r = [tex]\sqrt{49}[/tex]

r = 7 meters

Diameter of circle = 2 × (radius of the circle)

                              = 2 × 7

                              = 14 meters

Therefore, diameter of the circle is 14 meters.

Option (2) is the answer.

Answer:

14m

Step-by-step explanation:

Before Andrei adds the marbles to the glass tank, the glass tank was empty. This means that the volume of the empty tank is when n = 0 and the volume is 32 liters.

Given that:

[tex]W = 32 - 0.05n[/tex]

A linear function is represented as:

[tex]y = b + mx[/tex]

Where

[tex]b \to[/tex] y intercept

Literally, the y intercept is the initial value of the function.

In this function, the y intercept means the initial volume of the glass tank before filling it with marbles.

Compare [tex]y = b + mx[/tex] and [tex]W = 32 - 0.05n[/tex]

[tex]b = 32[/tex]

This means that the volume of the glass tank is 32 liters.

Read more about linear functions at:

https://brainly.com/question/21107621

Answer:

0.05

Step-by-step explanation:

Answer:

The number of the class containing the  largest score can be found in frequency  24 and the class is  98 - 103

For the third class 78 - 83 ; the upper limit = 83

The class width for this histogram 5

The number of students that took the exam simply refers to the frequency is  84

The percentage of students that scored higher than 95.5 is 53%

Step-by-step explanation:

The objective of this question is to use the following histogram that shows the exam scores for a Pre-algebra class to answer the question given:

NOW;

The table given in the question can be illustrated as follows:

S/N           Class                Score          Frequency

1                 68  - 73            70.5           0

2                73 - 78             75.5           4

3                78 - 83             80.5           8

4                83 - 88             85.5           12

5                88 - 93            90.5           16

6                93 - 98            95.5           20

7                98 - 103           100.5          24                            

TOTAL:                                                 84

a) The number of the class containing the  largest score can be found in frequency  24 and the class is  98 - 103

b) For the third class 78 - 83 ; the upper limit = 83  ( since the upper limit is derived by addition of  5 to the last number showing  in the highest value specified by the number in the class interval which is 78 ( i.e 78 + 5 = 83))

c)   The class width for this histogram 5 ; since it  is the difference  between the upper and lower boundaries  limit of the given class.

So , from above the difference in any of the class will definitely result into 5

d) The number of students that took the exam simply refers to the frequency ; which is (0+4+8+12+16+20+24) = 84

e) Lastly; the percentage of students that scored higher than 95.5 is ;

⇒[tex]\dfrac{20+24}{84} *100[/tex]

= 0.5238095 × 100

= 52.83

To the nearest percentage ;the percentage of students that scored higher than 95.5 is 53%

Answer:

1. 98-103 (6th class)

2. 88

3. 5

4. 84

5. 52%

Step-by-step explanation:

Find attached the frequency table.

The class of exam scores falls between (1, 2, 3, 4, 5, or 6).

The exam score ranged from 68-103

1) The largest number of exam scores = 24

The largest number of exam scores is in the 6th class = 98 -103

Step 2 of 5:

The upper class limit is the higher number in an interval. Third class interval is 83-88

The upper class limit of the third class 88.

Step 3 of 5:

Class width = upper class limit - lower class limit

We can use any of the class interval to find this as the answer will be the same. Using the interval between 73-78

Class width = 78 - 73

Class width for the histogram = 5

Step 4 of 5:

The total of students that took the test = sum of all the frequency

= 0+4+8+12+16+20+24 = 84

The total of students that took the test = 84

Step 5 of 5:Find the percentage of students that scored higher than 95.5

Number of student that scored higher than 95.5 = 20 + 24 = 44

Percentage of students that scored higher than 95.5 = [(Number of student that scored higher than 95.5)/(total number of students that took the test)] × 100

= (44/84) × 100 = 0.5238 × 100 = 52.38%

Percentage of students that scored higher than 95.5 = 52% (nearest percent)

Answer:

5-i

Step-by-step explanation:

Product=multiplication

Let the complex number=x

(3+2i)*x=17+7i

x=17+7i / 3+2i

x=(17-7i)*(3-2i)/(3+2i)*(3+2i)

=51-34i+21i+14i^2 / 9+6i+6i+4i^2

=51+13i+14i^2 / 9+12i+4i^2

= (51+14 - 13i) / 13

= (65 -13i) / 13

= 65 / 13 - 13 i / 13

= 5 - i.

Answer:

979

Step-by-step explanation:

Answer:

115/198

Step-by-step explanation:

khan

Answer:

a) Q1= 144.10

Median = 150

Q3=155.90

b) The 99 percentile would be:[tex]a=150 +2.33*8.75=170.39[/tex]

And represent a value who accumulate 99% of the values below

Step-by-step explanation:

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(150,8.75)[/tex]  

Where [tex]\mu=150[/tex] and [tex]\sigma=8.75[/tex]

Part a

Lets begin with the first quartile:

[tex]P(X>a)=0.75[/tex]   (a)

[tex]P(X<a)=0.25[/tex]   (b)

We can find the quantile in the normal standard distribution and we got z=-0.674.

And we can apply the z score formula and we got:

[tex]z=-0.674<\frac{a-150}{8.75}[/tex]

And if we solve for a we got

[tex]a=150 -0.674*8.75=144.10[/tex]

The median for this case is the mean [tex]Median =150[/tex]

For the third quartile we find the quantile who accumulate 0.75 of the area below and we got z=0.674 and we got:

[tex]a=150 +0.674*8.75=155.90[/tex]

Part b

We can find the quantile in the normal standard distribution who accumulate 0.99 of the area below and we got z=2.33.

And we can apply the z score formula and we got:

[tex]z=2.33<\frac{a-150}{8.75}[/tex]

And if we solve for a we got

[tex]a=150 +2.33*8.75=170.39[/tex]

And represent a value who accumulate 99% of the values below

Answer:

2 2/3

Step-by-step explanation:

Answer:

The capital B refers to the base of the area

Step-by-step explanation:

Answer:

A

Step-by-step explanation:

The capital B means the area of the base

Answer:

I think it’s -4

Step-by-step explanation:

-2 plus -2 is -4

since 2 plus 2 is 4

Hope this helps ;)

Answer:

Step-by-step explanation:

It is because 2 plus 2 is 4

so -2 plus -2 is -4

hth!