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Crank
3 years ago
14

Angles P and Q are supplementary. If mZP is 32°, what is the mZQ?

Mathematics
1 answer:
Ratling [72]3 years ago
6 0

Step-by-step explanation:

Supplementary angles add up to 180°

\therefore m\angle P + m\angle Q = 180° \\ \therefore 32 \degree+ m\angle Q = 180° \\ \therefore m\angle Q = 180° - 32° \\  \huge \purple{ \boxed{\therefore m\angle Q = 148°}}

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There are 14 students in an art class. Mr.
Alexus [3.1K]
You take the number of the colored pencils and divides it with the people in the class.

350/14 = 25
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3 years ago
The sum of the present ages of the mother and her daughter is 80 years. The mother was thrice as old as her daughter when she wa
Alika [10]

Answer:

let the daughter age be x

let mother age be 3x

Step-by-step explanation:

x+3x= 80

4x = 80

x= 80\4

x=20

daughter age is 20

mather age= 80 -20

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daughter age 20

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7 0
3 years ago
Find the area and the circumference of a circle with radius 6 m.
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Answer:

(a) 113.04m²

(b) 37.68m

Step-by-step explanation:

(a)

Area = πr²

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(b)

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6 0
3 years ago
Read 2 more answers
In a flower garden, there are 4 tulips for every 9 daisies. If there are 24 tulips, how many daisies are there? Please help i wi
miss Akunina [59]

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4 0
3 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
3 years ago
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