How do you solve the formula for d=rt for t

Factor the polynomial. 7x2 + 68xy - 20y2 A) (7x - 2y)(x + 10y) B) (7x - 10y)(x - 2y) C) (7x + 10y)(x + 2y) D) (7x + 2y)(x - 10y)

Otrada [13]

For this case we have the following polynomial:
7x2 + 68xy - 20y2
Factoring we have:
(7x-2y) (x + 10y)
We verify the factorization:
7x2 + 70xy - 2xy - 20y2
Rewriting we have:
7x2 + 68xy - 20y2
Therefore, the factorization is correct.
Answer:
A) (7x - 2y) (x + 10y)

5 0

2 years ago

What is 4.58 when approximated to the nearest hundredth

Ann [662]

Answer:

The answer would be 4.6

Step-by-step explanation:

4.58

8 is closest to 10 so round the 5 to 6.

6 0

2 years ago

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Five men and 5 women are ranked according to their scores on an examination. Assume that no two scores are alike and all possibl

serg [7]

Answer:

P(X=i) where i = 1,2,3,4,5,6,7,8,9,10 = 1/2, 5/18, 5/36, 5/84, 5/252, 1/252, 0, 0, 0, 0.

Step-by-step explanation:

X denotes the highest ranking achieved by the woman.

When X=1, the top ranked person is a female.

When X=2, the first person is a male and the second ranked person is a female.

Similarly, when X=3, the first two ranked persons are male and the third one is a female.

When X=4, the first three persons are male and the fourth one is a female.

When X=5, the first four persons are males and the fifth person is a female.

When X=6, the first five people are males and the sixth person is a female. The rest of the four people are also females since there are only five men in a sample space.

The probability for X=7, 8, 9, 10 is zero because there are only five men who can achieve the first five positions and the last highest rank that can be achieved by a woman is 6.

To compute the probabilities, we will use the formula:

No. of ways a female can be ranked X/Total number of ways to rank 10 people

Note that the total number of ways of ranking 10 different people is 10P10 or 10!

For X=1, the first position can be taken by any of the 5 women. The possible ways of the first person being a woman is 5C1. The rest of the 9 people can take any of the ranks. They can be ordered in 9P9 ways.

So, P(X=1) = (5C1)(9P9)/(10P10) = (5 x 362880)/(3628800) = 1/2

For X=2, the first rank must be taken by a male. The number of ways to arrange the first person as a male out of the 5 men can be calculated by 5P1. The second position must be taken by a female and rest of the 8 positions can be taken by any of the 8 people in 8P8 ways.

So, P(X=2) = (5P1)(5C1)(8P8)/(10P10) = (5 x 5 x 40320)/(3628800) = 5/18

For X=3, first two people must be men and the number of ways to arrange 2 out of 5 males at the first two positions is 5P2. The third position is a female. The rest of the 7 people can be ordered in 7P7 ways.

P(X=3) = (5P2)(5C1)(7P7)/(10P10) = (20 x 5 x 5040)/(3628800) = 5/36

P(X=4) = (5P3)(5C1)(6P6)/(10P10) = (60 x 5 x 720)/(3628800) = 5/84

P(X=5) = (5P4)(5C1)(5P5)/(10P10) = (120 x 5 x 120)/(3628800) = 5/252

P(X=6) = (5P5)(5C1)(4P4)/(10P10) = (120 x 5 x 24)/(3628800) = 1/252

P(X=7) = 0

P(X=8) = 0

P(X=9) = 0

P(X=10) = 0

7 0

3 years ago

A sample of 1500 computer chips revealed that 32% of the chips do not fail in the first 1000 hours of their use. The company's p

Grace [21]

Answer:

Yes, we have sufficient evidence at the 0.02 level to support the company's claim.

Step-by-step explanation:

We are given that a sample of 1500 computer chips revealed that 32% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature claimed that above 29% do not fail in the first 1000 hours of their use.

Let Null Hypothesis, $H_0$ : p $\leq$ 0.29 {means that less than or equal to 29% do not fail in the first 1000 hours of their use}

Alternate Hypothesis, $H_1$ : p > 0.29 {means that more than 29% do not fail in the first 1000 hours of their use}

The test statics that will be used here is One-sample proportions test;

T.S. = $\frac{\hat p -p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = proportion of chips that do not fail in the first 1000 hours of their use = 32%

n = sample of chips = 1500

So, test statistics = $\frac{0.32-0.29}{\sqrt{\frac{0.32(1-0.32)}{1500} } }$

= 2.491

Now, at 0.02 level of significance the z table gives critical value of 2.054. Since our test statistics is more than the critical value of z so we have sufficient evidence to reject null hypothesis as it fall in the rejection region.

Therefore, we conclude that more than 29% do not fail in the first 1000 hours of their use which means we have sufficient evidence at the 0.02 level to support the company's claim.

7 0

3 years ago

The Park Protectors is a group of elementary students and teachers from 32 schools who pick up trash at local parks. The line pl

just olya [345]

Answer:

~117

Step-by-step explanation:

3 3/4 of trash all together i divided it up by 3 3/4 / **32 x *1000

*1000=1 ton or 1000lbs

**32 the amount of schools

(i actually got 117.1875)

8 0

2 years ago

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