All categories

2 years ago

How many times will 12 go into 80

Mathematics

2 answers:

eduard2 years ago

6 0

$\frac{80}{12}$

$6 \frac{8}{12}$

$6\frac{2}{3}$

80 goes into 12 6 times with remainder of the number of 8

Burka [1]2 years ago

4 0

12 can go into 80, 6 times

You might be interested in

PLEASE HELP!!

Dima020 [189]

Answer:

A. 90°+m∠4

Step-by-step explanation:

Also

∠ A + ∠ C = 180 o = ∠ B + ∠ D

⇒ 2 x+ ( 3 x − 5 ) = 180 0 = ( x + 5 ) + ∠ C

5 x − 5 = 180

Add 5 to both sides

5 x = 185

Divide both sides by 5

x = 185 5 = 37

But it was A. 90°+m∠4

4 0

2 years ago

What is the answer of these, how do you do these

Charra [1.4K]

1 is perpendicular bc it's the opposite reciprocal so Basically it's the original fraction upside down. 2 is parallel bc both fractions have the same slope but one is negative

3 0

3 years ago

Write 6.2 as a fraction and in word form.

nevsk [136]

6.2:

fraction: .2 = 0.20 = 20/100

fraction: 6 20/100

Word form: six point two

two and five hundredths

fraction: 2 5/100

decimal form 2.05

hope this helps

3 0

3 years ago

Read 2 more answers

Please answer and explain.

almond37 [142]

Hi there!

10. the number: x

4 more = addition

5 times = multiplication

less then 54 = 54 >

4 + 5x = <54

Hope this helps!

4 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.

Let p = population proportion of users who develop nausea

So, Null Hypothesis, $H_0$ : p = 0.20

Alternate Hypothesis, $H_a$ : p > 0.20

Here, null hypothesis states that 20% of users develop nausea.

And alternate hypothesis states that more than 20% of users develop nausea.

The test statistics that would be used here is One-sample z proportion 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

3 years ago