Ask question

Ask question

All categories

Naddika [18.5K]

3 years ago

14

Sienna has a coupon for an additional 15% off her entire purchase at a store. She knows she can solve the expression p-0.15p to

find out her total cost. Which of the following expressions could Sienna also use to find her total cost?
A. 0.75p
B. 0.85p
C. 1.15p
D. 1.85p

1 answer:

snow_lady [41]3 years ago

5 0

It is B because 100% -15%=.85

You might be interested in

Please help! (Find the area of the shaded region)

jeyben [28]

Answer:

9.43 ft

Step-by-step explanation:

First, we need to find the area of the entire circle(shaded & unshaded), which is 12.57, with a radius of 2 ft. Next, we find the area of the white area, and NOT the shaded area, which is 3.14, with a radius of 1 ft. Now, we need to subtract the white area(3.14) from the total area(12.57), which is 9.43.

Hope this helps!

8 0

3 years ago

How many feet is 48 inches

Ivenika [448]

48 inches is equal to 4 feet

5 0

3 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

Please help with this qn :)

zepelin [54]

Answer:

4 is the length

Step-by-step explanation:

14 and 4

14 x 14 + 4 x 4 = 212

14 + 4 = 18

7 0

2 years ago

A container holds 4 gallons of lemonade. A large lemonade contains 16 ounces. How many large lemonades could the restaurant sell

s344n2d4d5 [400]

Answer:

32

Step-by-step explanation:

4 0

3 years ago