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3 years ago

15

A set of data with a mean of 45 and a standard deviation of 8.3 is normally distributed. Find the value that is –2 standard devi

ations away from the mean

1 answer:

Varvara68 [4.7K]3 years ago

5 0

I believe it is 34% + 13.5% = 47.5%
0.475(8.3)=3.9425
45-3.9425=41.0575

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On June 1st, Lucy & Bros received an order for 500 cupcakes. Lucy delivered the cupcakes to the client on June 25th. A $50 d

Verizon [17]

Answer:

The revenue will be recognized on the June 25th

Step-by-step explanation:

Data provided in the question:

Date on which the order for 500 cupcakes was received is <u>June 1st</u>

Date on which the order for 500 cupcakes was delivered is <u>June 25th</u>

Date on which the deposit of $50 was received is June 5th

Date on which the remaining $450 was received is June 30th

Now,

Revenue is always recognized as and when revenue generated and order completes.

Therefore,

In the given question, the order was delivered on June 25th

Hence,

The revenue will be recognized on the June 25th

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3 years ago

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

Korolek [52]

Answer:

A sample of 499 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is given by:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

In this question, we have that:

$\pi = 0.21$

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

How large a sample would be required in order to estimate the fraction of tenth graders reading at or below the eighth grade level at the 90% confidence level with an error of at most 0.03

We need a sample of n, which is found when $M = 0.03$ . So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.645\sqrt{\frac{0.21*0.79}{n}}$

$0.03\sqrt{n} = 1.645\sqrt{0.21*0.79}$

$\sqrt{n} = \frac{1.645\sqrt{0.21*0.79}}{0.03}$

$(\sqrt{n})^2 = (\frac{1.645\sqrt{0.21*0.79}}{0.03})^2$

$n = 498.81$

Rounding up

A sample of 499 is needed.

8 0

3 years ago

Between 0 degrees Celsius and 30 degrees Celsius, the volume V (in cubic centimeters) of 1 kg of water at a temperature T is giv

Ira Lisetskai [31]

Answer:

T = 3.967 C

Step-by-step explanation:

Density = mass / volume

Use the mass = 1kg and volume as the equation given V, we will come up with the following equation

D = 1 / 999.87−0.06426T+0.0085043T^2−0.0000679T^3

= (999.87−0.06426T+0.0085043T^2−0.0000679T^3)^-1

Find the first derivative of D with respect to temperature T

dD/dT = $\dfrac{70000000\left(291T^2-24298T+91800\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^2}$

Let dD/dT = 0 to find the critical value we will get

$\dfrac{70000000\left(291T^2-24298T+91800\right)}$ = 0

Using formula of quadratic, we get the roots:

T = 79.53 and T = 3.967

Since the temperature is only between 0 and 30, pick T = 3.967

Find 2nd derivative to check whether the equation will have maximum value:

$-\dfrac{140000000\left(395178T^4-65993368T^3+3286558821T^2+2886200857800T-121415215620000\right)}{\left(679T^3-85043T^2+642600T-9998700000\right)^3}$

Substituting the value with T=3.967,

d2D/dT2 = -1.54 x 10^(-8) a negative value. Hence It is a maximum value

Substitute T =3.967 into equation V, we get V = 0.001 i.e. the volume when the the density is the highest is at 0.001 m3 with density of

D = 1/0.001 = 1000 kg/m3

Therefore T = 3.967 C

3 0

4 years ago

The population of Stark county this year is 7,921 people. Last year the population was 7,547 people. Find the approximate percen

Sholpan [36]

Answer:

Option (C)

Step-by-step explanation:

To determine the population we use the formula,

$P_n=P_0(1+r)^{n}$

Where $P_n$ = final population

$P_0$ = Initial population

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Now we substitute these values in the formula,

$7921=7547(1+r)^1$

$\frac{7921}{7547}=(1+r)$

r = 1.0496 - 1

r = 0.0496

r = 0.05

Therefore, percentage growth of the population of Stark county is 5%.

Option (C). will be the answer.

4 0

4 years ago

Toby the cat had quite a day. Toby caught 14 mice and ate 9 of them. How many mice does Toby still HAVE TO eat?

schepotkina [342]

Answer:

5

Step-by-step explanation:

14-9 = 5

3 0

3 years ago

Read 2 more answers