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Tju [1.3M]
3 years ago
7

Michael wants to find the length of the shadow of a tree. He measures the height of a fencepost and the length of the shadow it

casts. The fencepost is 3.5 feet tall, and its shadow is 10.5 feet long. Next, Michael measures the height of the tree, and finds it is 6 feet tall. How long is the shadow of the tree? ______________________________________
Mathematics
1 answer:
Irina18 [472]3 years ago
6 0
10.5÷3.5 = 3
6x3 = 18
18 feet is your answer (hope this helps)
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Find the leading coefficient and write the factored form of the function of least degree for the polynomial in the graph below.
OLga [1]
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3 0
2 years ago
Solve for y.<br> -16y - 7y + -5 = 18<br> Y =
Anton [14]

Answer: y = - 1

Step-by-step explanation:

  1. Combine the like terms. -23y - 5 = 18
  2. Add 5 on both sides. -23y = 23
  3. Divide -23 on both sides. y = - 1
4 0
2 years ago
Read 2 more answers
a store is having a clearance sale where merchandise on the sales racks is reduced by 80% from the original. if a jacket was ori
posledela

The sales price is $15.2

Step-by-step explanation:

Given,

Discount = 80%

Original price = $76

Discount price = 80% of original price

Discount\ price=\frac{80}{100}*76\\Discount\ price=\frac{6080}{100}\\Discount\ price=\$60.8

Sale's price = Original price - Discount price

Sale\ price= 76-60.8\\Sale\ price= \$15.2

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Keywords: Subtraction, percentage

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8 0
3 years ago
Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65
erma4kov [3.2K]

Answer:

Probability that the sample average is at most 3.00 = 0.98030

Probability that the sample average is between 2.65 and 3.00 = 0.4803

Step-by-step explanation:

We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.85.

Also, a random sample of 25 specimens is selected.

Let X bar = Sample average sediment density

The z score probability distribution for sample average is given by;

               Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean = 2.65

           \sigma  = standard deviation = 0.85

            n = sample size = 25

(a) Probability that the sample average sediment density is at most 3.00 is given by = P( X bar <= 3.00)

    P(X bar <= 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 2.06) = 0.98030

(b) Probability that sample average sediment density is between 2.65 and 3.00 is given by = P(2.65 < X bar < 3.00) = P(X bar < 3) - P(X bar <= 2.65)

P(X bar < 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z < 2.06) = 0.98030

 P(X bar <= 2.65) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{2.65-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 0) = 0.5

Therefore, P(2.65 < X bar < 3)  = 0.98030 - 0.5 = 0.4803 .

                                                                             

8 0
3 years ago
One uranium atom has a mass of 3.95 x 10^-22 grams.
creativ13 [48]
Remember
(ab)/(cd)=(a/c)(b/d)
and
\frac{x^n}{x^m}=x^{n-m}

1kg=1000g=1*10^3g
hmm, I estimate, 4 of them

x=number of attoms
mass of 1 times x=1*10^3
3.95*10^-22 times x=1*10^3
divide both sides by 3.95*10^-22
x=\frac{1*10^3}{3.95*10^{-22}}= (\frac{1}{3.95}  )( \frac{10^{3}}{10^{-22}} )=0.25316*10^{25}=2.5316*10^{24}
a million has 9 zeroes
so 24 zeroes is alot that is a <span>septillion, so there are 2.5316 septillion uranium atoms in 1kg


a. 4
b. underestimate

</span>
5 0
3 years ago
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