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3 years ago

EXPERT HELP I'LL GIVE BRAINLIEST:

Mathematics

2 answers:

timurjin [86]3 years ago

7 0

Answer:

990, C

Step-by-step explanation:

The formula for the surface area of the bottom cushion is

A = 2wl + 2lh + 2hw (w is width, l is length, h is height)

w = 18, l = 21, h = 3

A = 2(18)(21) + 2(21)(3) + 2(3)(18)

A = 2(378) + 2(63) + 2(54)

A = 756 + 126 + 108

A = 990 (C)

makkiz [27]3 years ago

6 0

The answer is c. credits to the person above me. :]

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Simplify 2 x times the fraction 1 over x to the power of negative 3 times x to the power of negative 2.

Ahat [919]

Answer: 1 over x to the 5th power

Step-by-step explanation:

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3 years ago

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Two machines are used for filling glass bottles with a soft-drink beverage. The filling process have known standard deviations s

stellarik [79]

Answer:

a. We reject the null hypothesis at the significance level of 0.05

b. The p-value is zero for practical applications

c. (-0.0225, -0.0375)

Step-by-step explanation:

Let the bottles from machine 1 be the first population and the bottles from machine 2 be the second population.

Then we have $n_{1} = 25$ , $\bar{x}_{1} = 2.04$ , $\sigma_{1} = 0.010$ and $n_{2} = 20$ , $\bar{x}_{2} = 2.07$ , $\sigma_{2} = 0.015$ . The pooled estimate is given by

$\sigma_{p}^{2} = \frac{(n_{1}-1)\sigma_{1}^{2}+(n_{2}-1)\sigma_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(25-1)(0.010)^{2}+(20-1)(0.015)^{2}}{25+20-2} = 0.0001552$

a. We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} \neq 0$ (two-tailed alternative).

The test statistic is $T = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$ and the observed value is $t_{0} = \frac{2.04 - 2.07}{(0.01246)(0.3)} = -8.0257$ . T has a Student's t distribution with 20 + 25 - 2 = 43 df.

The rejection region is given by RR = {t | t < -2.0167 or t > 2.0167} where -2.0167 and 2.0167 are the 2.5th and 97.5th quantiles of the Student's t distribution with 43 df respectively. Because the observed value $t_{0}$ falls inside RR, we reject the null hypothesis at the significance level of 0.05

b. The p-value for this test is given by $2P(T$ 0 (4.359564e-10) because we have a two-tailed alternative. Here T has a t distribution with 43 df.

c. The 95% confidence interval for the true mean difference is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.05/2}s_{p}\sqrt{\frac{1}{25}+\frac{1}{20}}$ , i.e.,

$-0.03\pm t_{0.025}0.012459\sqrt{\frac{1}{25}+\frac{1}{20}}$

where $t_{0.025}$ is the 2.5th quantile of the t distribution with (25+20-2) = 43 degrees of freedom. So

$-0.03\pm(2.0167)(0.012459)(0.3)$ , i.e.,

(-0.0225, -0.0375)

8 0

3 years ago

A sum of 17 and a difference 3

sleet_krkn [62]

It would be 9-6=3 and 9+6=17 hope this help

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3 years ago

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I NEED HELP PLEASE ASAP !

In-s [12.5K]

The first rule you need to know is

$(a^b)^c=a^{bc}$

So, you have

$\dfrac{(x^{25})^{-6}}{(x^{-3})^{48}}=\dfrac{x^{25\cdot(-6)}}{x^{-3\cdot 48}}=\dfrac{x^{-150}}{x^{-144}}$

The second rule you need to know is

$\dfrac{a^b}{a^c}=a^{b-c}$

So, you have

$\dfrac{x^{-150}}{x^{-144}}=x^{-150-(-144)}=x^{-150+144}=x^{-6}$

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3 years ago

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The data below represent the weight, in kilograms, of 15 containers shipped from

Nikolay [14]

Answer:

(A) Maximum = 100 kg.

(B) Lower quartile = 55 kg.

(D) Upper quartile = 93 kg.

(E) Median = 81 kg.

Step-by-step explanation:

A box-plot, also known as a box and whisker plot is a method to demonstrate the distribution of a data-set based on the following 5 number summary,

Minimum (shown at the bottom of the chart)
First Quartile (shown by the bottom line of the box)
Median (or the second quartile) (shown as a line in the center of the box)
Third Quartile (shown by the top line of the box)
Maximum (shown at the top of the chart).

The data provided for the weight, in kilograms, of 15 containers shipped from a factory in one week is as follows:

S = {45, 51, 53, 55, 55, 65, 75, 81, 84, 87, 93, 93, 95, 96, 100}

The data is already arranged in ascending order.

The maximum value is:

Maximum = 100 kg.

The first quartile is the median value of the first half of the data.

The first half of the data is:

S₁ = {45, 51, 53, 55, 55, 65, 75}

The median of this data set is the middle value, i.e. 4th value.

So, the first quartile is:

Lower quartile = 55 kg.

The minimum value is:

Minimum = 45 kg.

The third quartile is the median value of the second half of the data.

The second half of the data is:

S₂ = {84, 87, 93, 93, 95, 96, 100}

The median of this data set is the middle value, i.e. 4th value.

So, the third quartile is:

Upper quartile = 93 kg.

The The median of the data set is the middle value, i.e. the 8th value.

Median = 81 kg.

6 0

3 years ago