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2 years ago

Which statement best describes the relationship between the two cylinders shown below? (Use 3.14 for π.)

Mathematics

1 answer:

8090 [49]2 years ago

5 0

Answer:

The volume are equal

Step-by-step explanation:

Because when you multiply them both they are still equal.

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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

How do you write the following linear equations in standard form ?

ANTONII [103]

Hi, Victoriasoto93p63zva!

To rewrite these equations in standard form you need to understand that standard form looks something like this Ax+By=C, now obviously there will be variations like Ax-By=C or -Ax+By=C, etc..., but for these expressions, they would look like this:
1. -8y + 0x = 37
2. 14x + 4y = -18
3. x + $\frac{3}{5}$ y = $\frac{4}{5}$ 
4. 13x + 0y = -25

-ASIAX Frequent Answerer

4 0

3 years ago

A piece of pipe is 303/4 inches long. If five pieces, each 41/3 inches long, are cut from the pipe, how many inches of pipe rema

galina1969 [7]

So, what is the total lenght of the pieces that wer cut off?

we multiply the lenght by the number:

4 $\frac{1}{3}$ *5=20+ $\frac{5}{3}$ =21 $\frac{2}{3}$

and we subtract this from the original pipe, that is

30 $\frac{3}{4}$ -21 $\frac{2}{3}$

we need to bring the two fractions to the same denominator (by multiplying the fraction art in the first by 3 and 4 the second):

30 $\frac{9}{12}$ -21 $\frac{8}{12}$ =9 $\frac{1}{12}$

so the correct answer is D!

3 0

3 years ago

The LCD for the fractions 1/3, 3/4, 5/32, and 8/9 is

Stells [14]

The LCD for the fractions 1/3, 3/4, 5/32, and 8/9 is
B. 288.

5 0

3 years ago

Can someone please help me?

shepuryov [24]

Answer:

Step-by-step explanation:

area = wl (6×4 = 24)

w width l length

4 0

2 years ago