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4 years ago

Which information is not necessary to solve this problem? WHO EVER GETS THIS FIRST IS BRAINLIEST

1 answer:

4 0

So we need to now how many crates they sold, how many boxes are in a crate, and how much they sold the cates for.

So we don't need A. we don't need to know how much is in the box, just how much money they made from the box.

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How to regroup to solve the subtraction problem what is the diffrence 316-226=

astra-53 [7]

316-226=90 90 is your answer

5 0

3 years ago

A 12-sided die is rolled. The set of equally likely outcomes is {1 comma 2 comma 3 comma 4 comma 5 comma 6 comma 7 comma 8 com

LenKa [72]

Answer:

Step-by-step explanation:

It is certain that the number rolled will be less than 14, so the probability is 1.

7 0

3 years ago

Read 2 more answers

One card is randomly selected from a deck of cards find the odds against drawing a seven

Stella [2.4K]

Answer:

$\frac{3}{4}$

Step-by-step explanation:

Probability of getting a 7 + Probability of not getting a 7 = 1.

Odds against drawing a 7 = Probability of not getting a 7.

We could calculate the probability of getting a 7 and subtract 1 from it to get the answer.

There a four sevens in a deck of card, one of each kind.

Probability of getting a 7 = $P(7) = \frac{4}{52} = \frac{1}{13}$

$\implies P(not \hspace{1mm} getting \hspace{1mm} a \hspace{1mm} 7) = 1 - P(7)$

$\implies$ P(not getting a 7) $= 1 - \frac{1}{13} = \frac{12}{13}$

4 0

4 years ago

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