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svetlana [45]
3 years ago
12

I don’t really get this question so if anyone knows it please help

Mathematics
1 answer:
Mars2501 [29]3 years ago
3 0

Answer:

116.82\:\text{square inches}

Step-by-step explanation:

The window consists of a large square and 4 smaller triangles on the outside. The side length of this square is marked as 10, and therefore the area of this square is 10^2=100\:\mathrm{in^2}.

The area of a triangle is given by \frac{1}{2}\cdot b\cdot h. Since the problem states that the overlapping squares are congruent, both legs of the triangle will be 2.9 and intersect at a 90^{\circ} angle.

Thus, the sum of all four triangles is \frac{1}{2}\cdot 2.9\cdot 2.9\cdot 4=16.82\:\mathrm{in^2}.

Therefore, the area of the window is:

100+16.82=\boxed{116.82\:\mathrm{in^2}}

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Answer:

q = -15

Step-by-step explanation:

5q = -75

Divide both sides by 5 : q = -15

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Henry drove 60.2 miles in 7 days how many miles did he drive in 1 day?
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Answer:

8.6

Step-by-step explanation:

60.2/7 =8.6

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A clinical test on humans of a new drug is normally done in three phases. Phase I is conducted with a relatively small number of
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A clinical test on humans of a new drug is normally done in three phases. Phase I is conducted with a relatively small number of healthy volunteers. For​ example, a phase I test of a specific drug involved only 7 subjects.

Step-by-step explanation:

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he amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and s
sp2606 [1]

Complete question:

He amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.3 minutes and standard deviation 1.4 minutes. Suppose that a random sample of n equals 47 customers is observed. Find the probability that the average time waiting in line for these customers is

a) less than 8 minutes

b) between 8 and 9 minutes

c) less than 7.5 minutes

Answer:

a) 0.0708

b) 0.9291

c) 0.0000

Step-by-step explanation:

Given:

n = 47

u = 8.3 mins

s.d = 1.4 mins

a) Less than 8 minutes:

P(X

P(X' < 8) = P(Z< - 1.47)

Using the normal distribution table:

NORMSDIST(-1.47)

= 0.0708

b) between 8 and 9 minutes:

P(8< X' <9) =[\frac{8-8.3}{1.4/ \sqrt{47}}< \frac{X'-u}{s.d/ \sqrt{n}} < \frac{9-8.3}{1.4/ \sqrt{47}}]

= P(-1.47 <Z< 6.366)

= P( Z< 6.366) - P(Z< -1.47)

Using normal distribution table,

NORMSDIST(6.366)-NORMSDIST(-1.47)

0.9999 - 0.0708

= 0.9291

c) Less than 7.5 minutes:

P(X'<7.5) = P [Z< \frac{7.5-8.3}{1.4/ \sqrt{47}}]

P(X' < 7.5) = P(Z< -3.92)

NORMSDIST (-3.92)

= 0.0000

3 0
2 years ago
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