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yawa3891 [41]
3 years ago
15

What distance in km would you have covered if you travelled at a speed of 72 km/H for 2 hours and 30 minuits? ​

Mathematics
1 answer:
algol133 years ago
5 0

Answer:180

Step-by-step explanation:

72 x 2 = 144

72 ÷ 2 = 36

144 + 36 = 180

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A store has a 28% off sale on hammers. Before the discount, the price of a hammer is $49. What is the sale price?
zheka24 [161]
28% of 49 is 13.72. So subtract that from 49 and you get 35.28. 35.28 is the price after the sale
5 0
2 years ago
Please can you answer this whole page please
sleet_krkn [62]

Answer:

.......................................................................................

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I need help!!! ASAP!!!! !!!!!
GalinKa [24]

Answer:

1.

Volume= 729cm^3

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Area of Base= 81cm^2

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Volume=450m^3

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Area of Base=150m^2

3.

Volume= 480 cm^2

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4.

Volume =120in^3

Area of Base=20in^2

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Step-by-step explanation:

Volume = (length) Times (Width) Times (Height)

Area os base = (Length) Times (width)

3 0
2 years ago
-8(2b-7)-9(b-5)<br>simplify ​
GenaCL600 [577]

Answer:

<h2>-25b + 101</h2>

Step-by-step explanation:

-8(2b-7)-9(b-5)\qquad\text{use the distributive property:}\ a(b+c)=ab+ac\\\\=(-8)(2b)+(-8)(-7)+(-9)(b)+(-9)(-5)\\\\=-16b+56-9b+45\qquad\text{combine like terms}\\\\=(-16b-9b)+(56+45)\\\\=-25b+101

3 0
3 years ago
Read 2 more answers
. Exam scores for a large introductory statistics class follow an approximate normal distribution with an average score of 56 an
Andru [333]

Answer:

0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 56, \sigma = 5, n = 20, s = \frac{5}{\sqrt{20}} = 1.12

What is the probability that the average score of a random sample of 20 students exceeds 59.5?

This is 1 subtracted by the pvalue of Z when X = 59.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{59.5 - 56}{1.12}

Z = 3.1

Z = 3.1 has a pvalue of 0.9990.

So there is a 1-0.9990 = 0.001 = 0.1% probability that the average score of a random sample of 20 students exceeds 59.5.

3 0
3 years ago
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