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r-ruslan [8.4K]
3 years ago
5

PLEASE ANSWER FAST I NEED ANSWER QUICK!!!

Mathematics
2 answers:
ser-zykov [4K]3 years ago
8 0
The answer is D. Hope it helps! :)
zhuklara [117]3 years ago
6 0
I agree with the other person, I also think that it's D
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A car rents for $250 per week plus $0.50 per mile.find the rental cost for a two-week trip of 500 miles for a group of three peo
Artyom0805 [142]
So 2 weeks mean
250x 2= 500
Then .50 per mile so 500 miles mean
.50 x 500= 250
In total it will be
500+250= 750
5 0
3 years ago
As an estimation, we are told £3 is €4.
cluponka [151]

Answer:22.78 to 2dp

Step-by-step explanation:

91.10 euros divide by 4 =22.775

=22.78 to 2dp

8 0
3 years ago
Bob has some 10 ltb weights and some 3 lb weights. Together, all his weights add up to 50 lbs. If x represents the number of 3 l
____ [38]

total weight = 50

3x + 10y = 50

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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, <u>test statistics</u> = \frac{0.32-0.29}{\sqrt{\frac{0.32(1-0.32)}{1500} } }

                              = 2.491

<em>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.</em>

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
To prepare a batch of cookies a factory uses 4/5 barrels of oatmeal in each batch. The factory used 4 4/5 barrels of oatmeal on
babymother [125]

Answer:

Step-by-step explanation:

4 4/5= 24/5  24/5\4/5=6/5= 1/15

6 0
3 years ago
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