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3 years ago

help: A train travels with a speed of 124 miles per hour. How much time would it take the train to travel 186 miles?

Mathematics

1 answer:

stealth61 [152]3 years ago

7 0

Answer:

23,064

Step-by-step explanation:

multiply two numbers

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Mara carried water bottles to the field to share with her team at halftime. The water bottles weighed a total of 60x2 + 48x + 24

ludmilkaskok [199]

For this case we have the following polynomial:

$60x ^ 2 + 48x + 24$

What we must do for this case is to factor the polynomial, so that we have:

1) The number of bottles

2) the weight of each bottle.

We have then:

$12 (5x ^ 2 + 4x + 2)$

Answer:

A factorization that could represent the number of water bottles and weight of each water bottle is:

$12 (5x ^ 2 + 4x + 2)$

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3 years ago

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Estimate the unit rate to the nearest hundredth. The unit rate is about $_per fluid ounce.

statuscvo [17]

Answer:

The unit rate is about $12per fluid ounce

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2 years ago

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Add -13/4+2/5 using the number line. select the location on the number line to plot the sum

Ivahew [28]

The answer is if you convert. we cant see the number line so, we cant tell you the second part of the problem.. attach the image please.

ok so convert.. -13/4 becomes -65/20 and 2/5 becomes 8/20, so add that and get -57/20, then simplify how you want to. thats your answer

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3 years ago

Anyone know how to do these? Show work if you do

Vlada [557]

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3 years ago

The probability that a male will be colorblind is .042. Find the probabilities that in a group of 53 men, the following are true

Dvinal [7]

Answer:

See below for answers and explanations

Step-by-step explanation:

Part A

$\displaystyle P(X=x)=\binom{n}{x}p^xq^{n-x}\\\\P(X=5)=\binom{53}{5}(0.042)^5(1-0.042)^{53-5}\\\\P(X=5)=\frac{53!}{(53-5)!*5!}(0.042)^5(0.958)^{48}\\\\P(X=5)\approx0.0478$

Part B

$P(X\leq5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)\\\\P(X\leq5)=\binom{53}{1}(0.042)^1(1-0.042)^{53-1}+\binom{53}{2}(0.042)^2(1-0.042)^{53-2}+\binom{53}{3}(0.042)^3(1-0.042)^{53-3}+\binom{53}{4}(0.042)^4(1-0.042)^{53-4}+\binom{53}{1}(0.042)^5(1-0.042)^{53-5}\\\\P(X\leq5)\approx0.9767$

Part C

$\displaystyle P(X\geq 1)=1-P(X=0)\\\\P(X\geq1)=1-(1-0.042)^{53}\\\\P(X\geq1)\approx1-0.1029\\\\P(X\geq1)\approx0.8971$

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2 years ago