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Tatiana [17]
3 years ago
9

The test scores for last week's History test were: 76, 100, 94, 90, 68, 90, 80, 92, 84 If test scores of 60 and 68 are added to

the list, what would happen to the mean and median? A) The mean and median stay the same. B) The mean and the median become equal. C) The mean and median get closer together. D) The mean and the median get further apart.Number of Sales Size 5 6 7 8 9 10 Number Sold 8 14 20 16 7 3 The data chart shows the number of sales for a particular style of shoe last month. Which measure of central tendency will help the store manager order the most popular size for restocking? A) mean B) median C) mode D) range
Mathematics
2 answers:
Licemer1 [7]3 years ago
5 0

Answer:

1. C

2.C mode is the size that is sold the most often

11111nata11111 [884]3 years ago
3 0

Answer:

Your answer for the first question is C

Step-by-step explanation:

This is true because you do the mean with the added numbers and without and you do the median with the added numbers and without

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Please need help on this
serious [3.7K]

4(x-2)=-72

Multiply the bracket by 4

(4)(x)(4)(-2)=-72

4x-8=-72

Move -8 to the other side. Sign changes from -8 to +8.

4x-8+8=-72+8

4x=-64

Divide by 4 for both sides.

4x/4=-64/4

x=-16

Answer : x=-16

6 0
4 years ago
Read 2 more answers
(2b^2 - 5b) - (7b + 3b^2)
Marta_Voda [28]

Answer:

-13b

Step-by-step explanation:

(2b^2 - 5b) - (7b + 3b^2)

-3b^2 - 10b^2 = -13b

8 0
4 years ago
A set of normally distributed data has a mean of 3.2 and a standard deviation of 0.7. Find the probability of randomly selecting
jenyasd209 [6]

Answer:

P(Z> 3.13) = 0.000874

Step-by-step explanation:

A set of normally distributed data has a mean of 3.2 and a standard deviation of 0.7. Find the probability of randomly selecting 30 values and obtaining an average greater than 3.6.

We can denote the population mean with the symbol \mu

According to the information given, the data have a population mean:

\mu = 3.2.

The standard deviation of the data is:

\sigma = 0.7.

Then, from the data, a sample of size n = 30 is taken.

We want to obtain the probability that the sample mean is greater than 3.6

If we call

\mu_m to the sample mean then, we seek to find:

P(\mu_m> 3.6)

To find this probability we find the Z statistic.

Z = \frac{\mu_m-\mu}{\sigma_{\mu_m}}

Where:

Where \sigma_{\mu_m} is the standard deviation of the sample

\sigma_{\mu_m} = \frac{\sigma}{\sqrt{n}}

\sigma_{\mu_m} =\frac{0.7}{\sqrt{30}}\\\\\sigma_{\mu_m} = 0.1278

P(\frac{\mu_m-\mu}{\sigma_{\mu_m}}> \frac{3.6-3.2}{0.1278})

Then:

Z = \frac{3.6-3.2}{0.1278}\\\\Z = 3.13

The probability sought is: P(Z> 3.13)

When looking in the standard normal probability tables for right tail we obtain:

P(Z> 3.13) = 0.000874

8 0
3 years ago
Find f. f ″(x) = x^−2, x > 0, f(1) = 0, f(6) = 0
marin [14]

If you do in fact mean f(1)=f(6)=0 (as opposed to one of these being the derivative of f at some point), then integrating twice gives

f''(x) = -\dfrac1{x^2}

f'(x) = \displaystyle -\int \frac{dx}{x^2} = \frac1x + C_1

f(x) = \displaystyle \int \left(\frac1x + C_1\right) \, dx = \ln|x| + C_1x + C_2

From the initial conditions, we find

f(1) = \ln|1| + C_1 + C_2 = 0 \implies C_1 + C_2 = 0

f(6) = \ln|6| + 6C_1 + C_2 = 0 \implies 6C_1 + C_2 = -\ln(6)

Eliminating C_2, we get

(C_1 + C_2) - (6C_1 + C_2) = 0 - (-\ln(6))

-5C_1 = \ln(6)

C_1 = -\dfrac{\ln(6)}5 = -\ln\left(\sqrt[5]{6}\right) \implies C_2 = \ln\left(\sqrt[5]{6}\right)

Then

\boxed{f(x) = \ln|x| - \ln\left(\sqrt[5]{6}\right)\,x + \ln\left(\sqrt[5]{6}\right)}

3 0
2 years ago
A random sample of 133 people was taken from a very large population. Sixty-five of the people in the sample were females.The st
rjkz [21]

Answer: 0.0433

Step-by-step explanation:

Given: Sample size : n= 133

The number of females in sample = 65

Then the proportion of females : P=\dfrac{65}{133}

The formula to calculate the standard error of the proportion is given by :-

S.E.=\sqrt{\dfrac{P(1-P)}{n}}

\Rightarrow S.E.=\sqrt{\dfrac{\dfrac{65}{133}(1-\dfrac{65}{133})}{133}}\\\\\Rightarrow\ \Rightarrow S.E.=0.0433444676341\approx0.0433

Hence, the standard error of the proportion of females is 0.0433.

6 0
3 years ago
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