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Simora [160]
3 years ago
14

ZDFG and LJKL are complementary angles. DFG = x + 1, and JKL = x - 7. Find the

Mathematics
1 answer:
Step2247 [10]3 years ago
7 0
I think the answer is D
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Is (9,5) a solution to this system of inequalities?
iren2701 [21]

Answer:

yes it is..............

8 0
2 years ago
a small dog’s heart beats about 530,000,000 times during its lifetime, and a large dog’s heart beats about 1.4 times this amount
iren2701 [21]

Answer:

I think the answer is 742,000,000

Step-by-step explanation:

6 0
3 years ago
The overhead reach distances of adult females are normally distributed with a mean of 205.5 and a standard deviation of 8.6 . a.
Anastasy [175]

Answer:

a) 0.073044

b) 0.75033

c) The normal distribution can be used in part (b), even though the sample size does not exceed 30 because initial population size is been distributed normally , therefore, the mean of the samples will be normally distributed regardless of their size(meaning whether the sample size is less than or equal to or exceeds 30, the sample means the mean of the samples will be normally distributed regardless of their size).

Step-by-step explanation:

The overhead reach distances of adult females are normally distributed with a mean of 205.5 and a standard deviation of 8.6 .

a. Find the probability that an individual distance is greater than 218.00 cm.

We solve using z score formula

z = (x-μ)/σ, where

x is the raw score = 218

μ is the population mean = 205.5

σ is the population standard deviation = 8.6

For x > 218

z = 218 - 205.5/8.6

z = 1.45349

Probability value from Z-Table:

P(x<218) = 0.92696

P(x>218) = 1 - P(x<218) = 0.073044

b. Find the probability that the mean for 15 randomly selected distances is greater than 204.00

When = random number of samples is given, we solve using this z score formula

z = (x-μ)/σ/√n

where

x is the raw score = 204

μ is the population mean = 205.5

σ is the population standard deviation = 8.6

n = 15

For x > 204

Hence

z = 204 - 205.5/8.6/√15

z = -0.67552

Probability value from Z-Table:

P(x<204) = 0.24967

P(x>204) = 1 - P(x<204) = 0.75033

c. Why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

The normal distribution can be used in part (b), even though the sample size does not exceed 30 because initial population size is been distributed normally , therefore, the mean of the samples will be normally distributed regardless of their size(meaning whether the sample size is less than or equal to or exceeds 30, the sample means the mean of the samples will be normally distributed regardless of their size).

3 0
3 years ago
An equilateral triangle has a base of 44 meters and a height of h meters. Inside that triangle is constructed a second equilater
TEA [102]

Answer:

6.25 %  of area of Large triangle =  area of smallest triangle

Step-by-step explanation:

Area of large triangle L = (1/2)*base*height

L = (1/2)*(44 m)* h

L = 22*h  square m.

2nd equilateral triangle area:  S =(1/2)*(22 m)*(0.5h)

S = 5.5*h sq. m.

3rd smallest equilateral triangle area : T = (1/2)*(11 m)*(0.25h)

T = 1.375*h sq. m

-----

Find percent P  where  T = P* L,     1.375*h = P * 22*h

P = 6.25%  = 0.0625

4 0
3 years ago
What is the answer -6x•-89-45g
Dennis_Churaev [7]
Factored
-3(-178x+15g)

Simplified
534x-45g
4 0
3 years ago
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