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lara [203]
3 years ago
9

What is the product? 3•[1 -4 5 -7]

Mathematics
2 answers:
krek1111 [17]3 years ago
4 0

Answer:

-153

Step-by-step explanation:

3-135-21

-132-21

-153

-BARSIC- [3]3 years ago
4 0

Answer:

I got -78

please i dont understand this

Hope that helps!! :)

Proably not though sorry

Bye bye have a great day <3

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A circle has a circumference of 7,850 units. What is the radius of the circle?
ioda

Answer:1250 units

Step-by-step explanation:

circumference=7850

Radius=?

Radius=circumference ➗ (2xπ)

Radius=7850 ➗ (2 x 3.14)

Radius=7850 ➗ (6.28)

Radius=1250 units

8 0
4 years ago
Plz help me out a bit :(<br> which one do you think it is?
Anna [14]
C. 9^4
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8 0
3 years ago
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If x= 5 and y= 2, then xy =
seropon [69]

Answer:

You cant find that out. X and Y are like two different animals in a zoo, and XY is a whole other animal, you cant combine two animals in the same cage, so X and Y cant turn into XY. You could do:

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8 0
3 years ago
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valkas [14]

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Step-by-step explanation:

8 0
2 years ago
Suppose 241 subjects are treated with a drug that is used to treat pain and 54 of them developed nausea. Use a 0.05 significance
lana66690 [7]

Answer:

Null Hypothesis, H_0 : p = 0.20  

Alternate Hypothesis, H_a : p > 0.20  

Step-by-step explanation:

We are given that 241 subjects are treated with a drug that is used to treat pain and 54 of them developed nausea.

We have to use a 0.05 significance level to test the claim that more than 20​% of users develop nausea.

<em>Let p = population proportion of users who develop nausea</em>

So, <u>Null Hypothesis,</u> H_0 : p = 0.20  

<u>Alternate Hypothesis</u>, H_a : p > 0.20  

Here, <u><em>null hypothesis</em></u> states that 20​% of users develop nausea.

And <u><em>alternate hypothesis</em></u> states that more than 20​% of users develop nausea.

The test statistics that would be used here is <u>One-sample z proportion</u> test statistics.

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

where,  \hat p = proportion of users who develop nausea in a sample of 241 subjects =  \frac{54}{241}  

             n = sample of subjects = 241

So, the above hypothesis would be appropriate to conduct the test.

6 0
4 years ago
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