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tino4ka555 [31]

3 years ago

8

The table shows the annual increase in cost in t years for certain grocery items. Based on the formula A = C(1 + r)t, which equa

tion could you use to determine the cost in t years for a gallon of milk?

2 answers:

muminat3 years ago

5 0

Given A=C(1+r)t
to get the formula for cost we make C the subject as follows
divide both sides by (1+r)t, this will give us:
A/(1+r)t=C
hence the cost function will be:
C=A(1+r)^(-t)

Andrei [34K]3 years ago

3 0

Answer:

The answers A = 4.25(1.10)^t

Step-by-step explanation:

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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

Information on a packet of seeds claims that the germination rate is 92%. The packet contains 160 seeds.

xenn [34]

Answer: a. N(0.92, 0.0215)

Step-by-step explanation:

For population proportion (p) and sample size n, the mean and standard deviation is given by :-

The sampling distribution model for p is given by :_

$N(\mu_p,\ \sigma_p)$

, where

$\mu_p=p \\ \sigma_p=\sqrt{\dfrac{p(1-p)}{n}$

We assume that the seeds are randomly selected.

Given : Information on a packet of seeds claims that the germination rate is 92%.

i.e. p= 0.92

The packet contains 160 seeds.

i.e. n= 160

Then , $\mu_p=0.92\\ \sigma_p=\sqrt{\dfrac{0.92(1-0.92)}{160}}\approx0.0215$

Hence, the sampling distribution model for p is:

$N(\mu_p,\sigma_p) = N(0.92, 0.0215)$

7 0

4 years ago