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3 years ago

7

Beverly drove from the Atlantic City to New York she drove 284 miles at a constant speed of 58 mph how long did it take Beverly

to complete the trip

1 answer:

lubasha [3.4K]3 years ago

8 0

Answer:

4.9 hours = 4 hours 54 minutes

Step-by-step explanation:

speed = distance/time

time * speed = distance

time = distance/speed

time = (284 miles)/(58 mph) = 4.9 hours

4.9 hours - 4 hours = 0.9 hours

0.9 hours * (60 minutes)/(1 hour) = 54 minutes

4.9 hours = 4 hours 54 minutes

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You need a mean of at least 90 points to advance to the next round of the touch-screen trivia. What score in the fifth game will

lianna [129]

Answer:

At least 98 is needed in the 5th game

Step-by-step explanation:

The missing parameters are:

$Game\ 1 = 95$

$Game\ 2 = 91$

$Game\ 3 = 77$

$Game\ 4 =89$

$Mean = 90$ at least

Required

The score in game 5 to make you advance

Mean is calculated as:

$Mean = \frac{\sum x}{n}$

So, we have:

$Mean = \frac{95 + 91 + 77 + 89 + Game\ 5}{5}$

$Mean = \frac{352 + Game\ 5}{5}$

The mean must be at least 90.

So, we have:

$\frac{352 + Game\ 5}{5} \ge 90$

Multiply both sides by 5

$5 * \frac{352 + Game\ 5}{5} \ge 90 * 5$

$352 + Game\ 5 \ge 450$

Make Game 5 the subject

$Game\ 5 \ge450 - 352$

$Game\ 5 \ge 98$

<em>At least 98 is needed in the 5th game</em>

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2 years ago

pam has 3 times as many green beads as red beads and twice as many blue beads as green beads. She has 150 beads in total. How ma

lilavasa [31]

25
$150 \div 3 =$
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3 years ago

If the function f(x)=mx+b has an inverse function, which statement must be true

skelet666 [1.2K]

Im pretty sure its
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2 years ago

Read 2 more answers

A graduate class consists of six students. What is the probability that at least 5 of them are born either in April or in Octobe

nlexa [21]

Answer:

0.0671% probability that at least 5 of them are born either in April or in October

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they were born in April or October, or they were not. The probabilty of a student being born in April or October is independent of other students. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Probability of a student being born in April or October

April has 30 days, October 31

The year has 365 days. So

$p = \frac{61}{365} = 0.167$

A graduate class consists of six students.

This means that $n = 6$

What is the probability that at least 5 of them are born either in April or in October?

$P(X \geq 5) = P(X = 5) + P(X = 6)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{6,5}.(0.167)^{5}.(0.833)^{1} = 0.000649$

$P(X = 6) = C_{6,6}.(0.167)^{6}.(0.833)^{0} = 0.000022$

$P(X \geq 5) = P(X = 5) + P(X = 6) = 0.000649 + 0.000022 = 0.000671$

0.0671% probability that at least 5 of them are born either in April or in October

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Given V=1/3(3.14)r^2h solve for r

Gnoma [55]

The answer is r = √ v/1.046667h (v over 1.046667h)

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3 years ago