Ask question

Ask question

All categories

gayaneshka [121]

2 years ago

14

A factory worker fills each container with 25 boxes of raisins. Each box has 1.33 ounces of raisins. If you buy 12 containers, h

ow many ounces of raisins will you buy?

1 answer:

marshall27 [118]2 years ago

5 0

You would buy 399 ounces of raisins

You might be interested in

One True Love? A survey that asked whether people agree or disagree with the statement ‘‘There is only one true love for each pe

sertanlavr [38]

Answer:

99% confidence interval for the proportion of people who disagree with the statement is [0.667 , 0.713].

Step-by-step explanation:

We are given that a survey that asked whether people agree or disagree with the statement ‘‘There is only one true love for each person." has been conducted. The result is that 735 of the 2625 respondents agreed, 1812 disagreed, and 78 answered ‘‘don’t know."

Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of people who disagree with the statement = $\frac{1812}{2625}$ = 0.69

n = sample of respondents = 2625

p = population proportion of people who disagree with statement

<em>Here for constructing 99% confidence interval we have used One-sample z proportion statistics.</em>

<u>So, 99% confidence interval for the population proportion, p is ;</u>

P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.58 & 2.58}

P(-2.58 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.58) = 0.99

P( $-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

<u>99% confidence interval for p</u> = [ $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.69-2.58 \times {\sqrt{\frac{0.69(1-0.69)}{2625} } }$ , $0.69+2.58 \times {\sqrt{\frac{0.69(1-0.69)}{2625} } }$ ]

= [0.667 , 0.713]

Therefore, 99% confidence interval for the proportion of people who disagree with the statement is [0.667 , 0.713].

5 0

3 years ago

Please answer this im going to fail Rewrite the fraction as a decimal.

ivolga24 [154]

Step-by-step explanation:

it ia 76.92 is the percentage for 10/13

8 0

3 years ago

An open-top box with a square base is to have a volume of 4 cubic feet. Find the dimensions of the box (in ft) that can be made

Lemur [1.5K]

Answer:

The dimension of box= $2ft\times 2ft\times 1ft$

Step-by-step explanation:

We are given that

Volume of box=4 cubic feet

Let x be the side of square base and h be the height of box

Volume of box= $lbh=x^2h$

$4=x^2h$

$h=\frac{4}{x^2}$

Now, surface area of box,A= $x^2+4xh$

$A=x^2+4x(\frac{4}{x^2})=x^2+\frac{16}{x}$

$\frac{dA}{dx}=2x-\frac{16}{x^2}$

$\frac{dA}{dx}=0$

$2x-\frac{16}{x^2}=0$

$2x=\frac{16}{x^2}$

$x^3=8$

$x=2$

$\frac{d^2A}{dx^2}=2+\frac{32}{x^3}$

Substitute x=2

$\frac{d^2A}{dx^2}=2+\frac{32}{2^3}=2+4=6>0$

Hence, the area of box is minimum at x=2

Therefore, side of square base,x=2 ft

Height of box,h= $\frac{4}{2^2}=1 ft$

Hence, the dimension of box= $2ft\times 2ft\times 1ft$

5 0

2 years ago

What is the mean for the following set of data?<br> 2,5, 7, 8, 11, 11, 12<br> 10<br> 11<br> 8

Yuri [45]

Add up all the numbers & divide by the number of numbers. That should give you the mean.

4 0

3 years ago

Read 2 more answers

Men are believed to have a slightly longer average length of short hospital stays than women; 5.2 days versus 4.5 days respectiv

Georgia [21]

Answer:

Since our calculated value is lower than our critical value, $z_{calc}=2.02$ , we have enough evidence to FAIL to reject the null hypothesis at 1% of singificance. So there is not nough evidence to conclude that mean for men it's significantly higher than the mean for female.

Step-by-step explanation:

1) Data given and notation

$\bar X_{1}=5.2$ represent the mean for group men

$\bar X_{2}=4.5$ represent the mean for group women

Assuming these values for the remaining data:

$\sigma_{1}=1.2$ represent the population standard deviation for the sample men

$\sigma_{2}=1.5$ represent the population standard deviation for the sample women

$n_{1}=32$ sample size for the group men

$n_{2}=30$ sample size for the group women

z would represent the statistic (variable of interest)

$p_v$ represent the p value

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for men it's higher than the mean for women, the system of hypothesis would be:

H0: $\mu_{1} \leq \mu_{2}$

H1: $\mu_{1} > \mu_{2}$

If we analyze the size for the samples both are higher than 30, and we know the population deviation's, so for this case is better apply a z test to compare means, and the statistic is given by:

$z=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

3) Calculate the statistic

We have all in order to replace in formula (1) like this:

$z=\frac{5.2-4.5}{\sqrt{\frac{1.2^2}{32}+\frac{1.5^2}{30}}}=2.02$

4) Find the critical value

In order to find the critical value we need to take in count that we are conducting a right tailed test, so we are looking on the normal standard distribution a value that accumulats 0.01 of the area on the right and 0.99 of the area on the left. We can us excel or a table to find it, for example the code in Excel is:

"=NORM.INV(1-0.01,0,1)", and we got $z_{critical}=2.33$

5) Statistical decision

Since our calculated value is lower than our critical value, $z_{calc}=2.02$ , we have enough evidence to FAIL to reject the null hypothesis at 1% of significance. So there is not nough evidence to conclude that mean for men it's significantly higher than the mean for female.

6 0

3 years ago