3 years ago

Really help me please

Mathematics

1 answer:

Ivanshal [37]3 years ago

6 0

144 x 3= (D) 432

Hope this can help! :)

You might be interested in

Evaluate 5x – 2y + (7x – y) for x = 7 and y = –2

Lynna [10]

Answer:

Step-by-step explanation:

5x – 2y + (7x – y)

Combine like terms

12x -3y

Let x = 7 and y = -2

12(7) -3(-2)

84 +6

6 0

3 years ago

Read 2 more answers

Factor: x^2 + 9x - 36 = 0

alexdok [17]

Answer:

I'm pretty sure it's x = 12, -3

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Help, quick! Worth 100 points

finlep [7]

Answer:

the answer is (C) 2(1-2x)=x-3

4 0

3 years ago

Find the average. 11,16,15,22 A.64 B.15 C.17 D.16

Artemon [7]

Your answer is D)16. To find the average, add up all the numbers and divivde by the number of numbers there is

7 0

3 years ago

Read 2 more answers

A data set includes 103 body temperatures of healthy adult humans having a mean of 98.3degreesF and a standard deviation of 0.73

faust18 [17]

Answer:

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher

Step-by-step explanation:

Data provided in the question:

sample size, n = 103

Mean temperature, μ = 98.3

Standard deviation, σ = 0.73

Degrees of freedom, df = n - 1 = 102

Now,

For Confidence level of 99%, and df = 102, the t-value = 2.62 [from the standard t table]

Therefore,

CI = $(Mean - \frac{t\times\sigma}{\sqrt{n}},Mean + \frac{t\times\sigma}{\sqrt{n}})$

Thus,

Lower limit of CI = $(Mean - \frac{t\times\sigma}{\sqrt{n}})$

Lower limit of CI = $(98.3 - \frac{2.62\times0.73}{\sqrt{103}})$

Lower limit of CI = 98.11

and,

Upper limit of CI = $(Mean + \frac{t\times\sigma}{\sqrt{n}})$

Upper limit of CI = $(98.3 + \frac{2.62\times0.73}{\sqrt{103}})$

Upper limit of CI = 98.49

Hence,

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher and the mean temperature could very possibly be 98.6°F

7 0

3 years ago