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natka813 [3]
3 years ago
6

9. Find the perimeter of this figure.​

Mathematics
1 answer:
goldenfox [79]3 years ago
3 0
112ft is the correct answer
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3 3/5 + 6 3/10 please
Dafna11 [192]
In order to solve first convert both values into improper fractions:
18/5+63/10
second, convert both values to fractions with a base of ten:
36/10+63/10
finally, add the numerators:
99/10 is the final answer

3 0
4 years ago
What is 1/3 as a decimal value?
Natalka [10]


.33 with a line over the .33 because it is not able to be simplified into a whole decimal when ever both numbers are a factor of 3 besides when it is half that will happen
4 0
3 years ago
Simplify <br> 3 (2x+4y-2z)+7 (x+y-4z)
Dima020 [189]
6x+12y-6z+7x7y-28z= 13x+19y-34z
6 0
3 years ago
A Simple random sample of 100 8th graders at a large suburban middle school indicated that 84% of them are involved with some ty
Setler [38]

Answer: a) (0.755, 0.925)

Step-by-step explanation:

Let p be the population proportion of 8th graders are involved with some type of after school activity.

As per given , we have

n= 100

sample proportion: \hat{p}=0.84

Significance level : \alpha= 1-0.98=0.02

Critical z-value : z_{\alpha/2}=2.33  (using z-value table)

Then, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity will be :-

\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

i.e. 0.84\pm (2.33)\sqrt{\dfrac{0.84(1-0.84)}{100}}

i.e. \approx0.84\pm 0.085

i.e. (0.84- 0.085,\ 0.84+ 0.085)=(0.755,\ 0.925)

Hence, the 98% confidence interval that estimates the proportion of them that are involved in an after school activity : a) (0.755, 0.925)

3 0
4 years ago
5/7 times 4/5=? in simplest form
lys-0071 [83]
5/7 * 4/5 = 20/35 = 4/7
7 0
3 years ago
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