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KiRa [710]

3 years ago

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Mathematics

1 answer:

Karo-lina-s [1.5K]3 years ago

3 0

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X − 0.5(k − 3x) = 1.5x − k In terms of k, what does x equal?

d1i1m1o1n [39]

104())(;34)-?-?!-'mhhm

7 0

2 years ago

Which expression is equivalent to (18x + 9k) + (2x + 6k)?

Marrrta [24]

Answer:

B. 20x + 15

Step-by-step explanation:

Get rid of the parenthesis and combine like terms: 18x + 2x and 9k + 6k

You get: 20x + 15.

5 0

3 years ago

Read 2 more answers

A student researcher compares the heights of men and women from the student body of a certain college in order to estimate the d

maria [59]

Answer:

(0.6613, 8.3387)

Step-by-step explanation:

Let the heights of men be the first population and the heights of women be the second population. Then

We have $n_{1} = 15$ , $\bar{x}_{1} = 69.4$ , $s_{1} = 3.09$ and

$n_{2} = 7$ , $\bar{x}_{2} = 64.9$ , $s_{2} = 2.58$ . The pooled

estimate is given by

$s_{p}^{2} = \frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(15-1)(3.09)^{2}+(7-1)(2.58)^{2}}{15+7-2} = 8.68$

The 99% confidence interval for the true mean difference between the mean height of the men and the mean height of the women is given by (if the samples are independent)

$(\bar{x}_{1}-\bar{x}_{2})\pm t_{0.01/2}s_{p}\sqrt{\frac{1}{15}+\frac{1}{7}}$ , i.e.,

$(69.4-64.9)\pm t_{0.005}2.946\sqrt{\frac{1}{15}+\frac{1}{7}}$

where $t_{0.005}$ is the 0.5th quantile of the t distribution with (15+7-2) = 20 degrees of freedom. So

$4.5\pm(-2.845)(2.946)(0.458)$ , i.e.,

(0.6613, 8.3387)

5 0

3 years ago

Simplify the expression (4 + 5i)(4 - 5i).<br>16 - 20i<br>16 + 20i<br>-9<br>41

AnnZ [28]

Answer:

The answer i got was 41. i just took a test with the same problem and it was correct. Hope this helps.

Step-by-step explanation:

3 0

2 years ago

There are three circular tabletops with diameters of 5 ft, 7 ft, and 9 ft. Identify the area of each table top rounded to the ne

ella [17]

It is almost about 19.6; 38.5; 63.6

6 0

2 years ago