Ask question

Ask question

All categories

3 years ago

11

Shen needs 271 programs for the school play on Thursday. How many boxes of programs will he need, given that each box contains 3

5 programs?

1 answer:

lesantik [10]3 years ago

8 0

7 because if you divide it by 35, it's seven.

You might be interested in

Is ( − 3,2 ) a solution of -4x - 10y < -9?<br><br>Yes or no

yanalaym [24]

Replace x and y in the equation with the given point and solve:

-4(-3) - 10(2) = 12 -20 = -8

-8 is not less than -9

(-3,2) is not a solution.

4 0

3 years ago

Read 2 more answers

Measurements of the sodium content in samples of two brands of chocolate bar yield the following results (in grams):

Tpy6a [65]

Answer:

98% confidence interval for the difference μX−μY = [ 0.697 , 7.303 ] .

Step-by-step explanation:

We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;

Brand A : 34.36, 31.26, 37.36, 28.52, 33.14, 32.74, 34.34, 34.33, 29.95

Brand B : 41.08, 38.22, 39.59, 38.82, 36.24, 37.73, 35.03, 39.22, 34.13, 34.33, 34.98, 29.64, 40.60

Also, $\mu_X$ represent the population mean for Brand B and let $\mu_Y$ represent the population mean for Brand A.

Since, we know nothing about the population standard deviation so the pivotal quantity used here for finding confidence interval is;

P.Q. = $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t_n__1+n_2-2$

where, $Xbar$ = Sample mean for Brand B data = 36.9

$Ybar$ = Sample mean for Brand A data = 32.9

$n_1$ = Sample size for Brand B data = 13

$n_2$ = Sample size for Brand A data = 9

$s_p = \sqrt{\frac{(n_1-1)s_X^{2}+(n_2-1)s_Y^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(13-1)*10.4+(9-1)*7.1 }{13+9-2} }$ = 3.013

Here, $s^{2}_X$ and $s^{2} _Y$ are sample variance of Brand B and Brand A data respectively.

So, 98% confidence interval for the difference μX−μY is given by;

P(-2.528 < $t_2_0$ < 2.528) = 0.98

P(-2.528 < $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 2.528) = 0.98

P(-2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(Xbar -Ybar) -(\mu_X-\mu_Y)$ < 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98

P( (Xbar - Ybar) - 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(\mu_X-\mu_Y)$ < (Xbar - Ybar) + 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98

98% Confidence interval for μX−μY =

[ (Xbar - Ybar) - 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ , (Xbar - Ybar) + 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ]

[ $(36.9 - 32.9)-2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}$ , $(36.9 - 32.9)+2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}$ ]

[ 0.697 , 7.303 ]

Therefore, 98% confidence interval for the difference μX−μY is [ 0.697 , 7.303 ] .

4 0

3 years ago

PLZ HELP WILL MARK BRAINLIEST THANKS

Anon25 [30]

90 I’m sorry if I am wrong I got that in paper :)

8 0

2 years ago

For the population of all college students currently taking a statistics course, you want to estimate the proportion who are wom

inessss [21]

Answer:

No

Step-by-step explanation:

To estimate population proportion from a sample, we must ensure that the sample data is random. Though a simple random sample of college students from a particular college was used as the sample data. However, selection of the college should have been randomized as well, a stratified random sample would have been a better sampling method whereby certain colleges are selected based on region or other criteria and then a random sample of it's statistics students selected. The sample proportion Obtian from a sample of this nature will be more representative of the population proportion of all college statistic student.

5 0

3 years ago

Richard deposits $237.95 every month into his mortgage. At the end of 30 years, he has a balance of $183,710.77. What interest h

77julia77 [94]

Answer:

$98,048.77

Step-by-step explanation:

First you want to find out how much he has put in without interest so you would do 237.95*12 to figure out how much he puts in per year then times that number by 30 to figure out how much he has put in in total, after this you subtract this total from the 183,710.77 to get the total amount of interest

3 0

3 years ago