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2 years ago

7th grade math for easy points

Mathematics

1 answer:

dedylja [7]2 years ago

7 0

Answer:

Step-by-step explanation:

8*8*8 = 512

8³ = 512

just punch it in a calculator

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How long would it take to travel 425 m at the rate of 50

Murljashka [212]

425 / 50 = 8,5

There you go.

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3 years ago

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Jason is 6 feet tall, and at 6 pm, his shadow was 15 feet long. At the same time, a tree next to Jason had a 25-foot shadow. Wha

IRISSAK [1]

Answer:

Step-by-step explanation:

set up a ratio problem

6 feet/15 long shad= x/25 shad

15x = 150

x = 10

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2 years ago

Please help me <br><br>If 180°<α<270°, cos⁡ α=−8/17, what is sin -α?

rewona [7]

Starting from the fundamental trigonometric equation, we have

$\cos^2(\alpha)+\sin^2(\alpha)=1 \iff \sin(\alpha)=\pm\sqrt{1-\cos^2(\alpha)}$

Since $180$ , we know that the angle lies in the third quadrant, where both sine and cosine are negative. So, in this specific case, we have

$\sin(\alpha)=-\sqrt{1-\cos^2(\alpha)}$

Plugging the numbers, we have

$\sin(\alpha)=-\sqrt{1-\dfrac{64}{289}}=-\sqrt{\dfrac{225}{289}}=-\dfrac{15}{17}$

Now, just recall that

$\sin(-\alpha)=-\sin(\alpha)$

to deduce

$\sin(-\alpha)=-\sin(\alpha)=-\left(-\dfrac{15}{17}\right)=\dfrac{15}{17}$

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2 years ago

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Determine the constant of variation for the direct variation given.

Flura [38]

The formula is y= k x where k is the constant of variation, so plug a coordinate in the formula to find value of k, lets get (3,12) you have 12 = 3k, then k = 4, if you get (9,36) then you have 36 = 9k where k = 4

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2 years ago

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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