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den301095 [7]
3 years ago
10

Someone please help :(!!!!!!

Mathematics
1 answer:
dezoksy [38]3 years ago
6 0

Answer:

GE

Step-by-step explanation:

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A concert cleanup crew has 6 workers and one cleanup kit.
myrzilka [38]

Answer:

Option (D)

Step-by-step explanation:

At every 6 crew workers number of clean up kits required = 1

Therefore, for 1 crew worker number of kits required = \frac{1}{6}

And number of kits required for w workers = \frac{w}{6} = w ÷ 6

If w = 54

Number of kits required = \frac{54}{6}=9

If w = 60

Number of kits required = \frac{60}{6}=10

If w = 66

Number of kits required = \frac{66}{6}=11

Therefore, table given in Option (D) is the correct table.

6 0
3 years ago
What is the length of segment AB?<br> Consider the diagram.<br> 7,9,18,25
Mrrafil [7]

Answer:

9

Step-by-step explanation:

It is congruent to the other side.

5 0
3 years ago
Can someone help me with this
melisa1 [442]
First part
option 2 p=KB

second part
p=$155,000k

or we can also find the value of k first
then the answer will be different :/
7 0
2 years ago
Convert 1012 into bainary number system​
rewona [7]
1111110100

Do modulo 2 then floor divide by 2 and repeat


Mark brainliest please
8 0
2 years ago
A publisher reports that 344% of their readers own a particular make of car. A marketing executive wants to test the claim that
inysia [295]

Answer:

No, there is not enough evidence at the 0.02 level to support the executive's claim.

Step-by-step explanation:

We are given that a publisher reports that 34% of their readers own a particular make of car. A random sample of 220 found that 30% of the readers owned a particular make of car.

And, a marketing executive wants to test the claim that the percentage is actually different from the reported percentage, i.e;

Null Hypothesis, H_0 : p = 0.34 {means that the percentage of readers who own a particular make of car is same as reported 34%}

Alternate Hypothesis, H_1 : p \neq 0.34 {means that the percentage of readers who own a particular make of car is different from the reported 34%}

The test statistics we will use here is;

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

where, p = actual % of readers who own a particular make of car = 0.34

            \hat p = percentage of readers who own a particular make of car in a

                  sample of 220 = 0.30

            n = sample size = 220

So, Test statistics = \frac{0.30 -0.34}{\sqrt{\frac{0.30(1- 0.30)}{220} } }

                             = -1.30

Now, at 0.02 significance level, the z table gives critical value of -2.3263 to 2.3263. Since our test statistics lie in the range of critical values which means it doesn't lie in the rejection region, so we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that the actual percentage of readers who own a particular make of car is same as reported percentage and the executive's claim that it is different is not supported.

3 0
3 years ago
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