Find two numbers whose sum is -2 and whose difference is -8

Solve the following proportion.<br> х<br> 42<br> 35<br> 10

anzhelika [568]

Answer:

12

Step-by-step explanation:

The proportion is equal to $\frac{x}{10}=\frac{42}{35}$

We can simplify the right side by dividing by 7/7 to get the right side equal to 6/5. Now we can cross-multiply to get

$60=5x$

Dividing both sides by 5, we have the answer as

$x=12$

6 0

3 years ago

The graph shows the depth, y, in meters, of a shark from the surface of an ocean for a certain amount of time, x, in minutes:

finlep [7]

Answer:

lo siento pero soy principiante

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

What is the length of side AB?

soldi70 [24.7K]

Answer:

6

Step-by-step explanation:

i have had this question before and got it right

4 0

4 years ago

Read 2 more answers

The true average diameter of ball bearings of a certain type is supposed to be 0.5 in. A one-sample t test will be carried out t

yulyashka [42]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

C

b

C

c

C

d

A

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 0.5 \ in$

Generally the Null hypothesis is $H_o : \mu = 0. 5 \ in$

The Alternative hypothesis is $H_a : \mu \ne 0.5 \ in$

Considering the parameter given for part a

The sample size is n = 15

The test statistics is t = 1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = 2.145$

We are making use of this $t_{\frac{\alpha }{2} }$ because it is a one-tail test

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that $t < t_{\frac{\alpha }{2} }$ so the null hypothesis would not be rejected

Considering the parameter given for part b

The sample size is n = 15

The test statistics is t = -1.66

The level of significance $\alpha = 0.05$

The degree of freedom is evaluated as

$df = n- 1$

$df = 15- 1$

$df = 14$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.05 }{2} ,14} = -2.145$

Looking at the value of t and $t_{\frac{\alpha}{2} ,df }$ the we see that t does not lie in the area covered by $t_{\frac{\alpha}{2} , df }$ (i.e the area from -2.145 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part c

The sample size is n = 26

The test statistics is t = -2.55

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = 2.787$

Looking at the value of t and $t_{\frac{\alpha }{2} }$ the we see that t does not lie in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we fail to reject the null hypothesis

Considering the parameter given for part d

The sample size is n = 26

The test statistics is t = -3.95

The level of significance $\alpha = 0.01$

The degree of freedom is evaluated as

$df = n- 1$

$df = 26- 1$

$df = 25$

Using the critical value calculator at (social science statistics web site )

$t_{\frac{\alpha}{2} ,df } = t_{\frac{0.01 }{2} ,25} = -2.787$

Looking at the value of t and $t_{\frac{\alpha}{2} }$ the we see that t lies in the area covered by $t_{\alpha , df }$ (i.e the area from -2.787 downwards on the normal distribution curve ) hence we reject the null hypothesis

6 0

3 years ago

The line plot (shown in the attached image) shows the prices of sunglasses at a department store.

Novosadov [1.4K]

a. Mean is 70.8, median is 70, and mode is 60

b. Mean describes the data best

c. Mean can be a misleading central measure in case of outliers

Calculating Mean, Median and Mode

Mean = Total sum of pries / Total number of sunglasses

Mean = (20 + 20 + 50 + 50 + 50 + 60 + 60 + 60 + 60 + 60 + 60 + 70 + 70 + 70 + 80 + 80 + 80 + 80 + 90 + 90 + 90 + 90 + 100 + 100 + 130) / 25

Mean = 70.8

Median = (n/2 + 1)th term

Here, n is the number of sunglasses.

Median = (25/2 + 1)th term

Median =13th term

Median = 70

Mode = Most occurring data

Mode = 70

Why is mean the best central measure to describe the data?

When your data distribution is continuous and symmetrical, such as when your data are normally distributed, the mean is typically the best measure of central tendency to utilize. Thus, mean describes the average price of the sunglasses here

Mean Being Misleading

In case there are outliers, the mean of the data gets highly affected. Hence, it can be misleading for analyzing average price of the data.

Learn more about mean here:

brainly.com/question/13451489

#SPJ1

7 0

2 years ago