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3 years ago

A truck driver drove 300 miles in 6 3/4 hours. How many miles per hour did the driver drive?

Mathematics

2 answers:

Sergio [31]3 years ago

5 0

Answer:

44.44 miles per hour.

Step-by-step explanation:

A truck driver drove 300 miles in $6\frac{3}{4}$ hours.

We have to calculate the speed per hour.

First we convert $6\frac{3}{4}$ hours to decimal form.

$6\frac{3}{4}$ = 6.75 hours

∵ in 6.75 hours the driver drove = 300 miles

∴ in 1 hour he drove = $\frac{300}{6.75}$ miles

= 44.44 miles

The truck driver drove 44.44 miles per hour.

Andru [333]3 years ago

3 0

44.4 miles per hour. You just divide 300 and 6.75 (Think as the 3/4 as quarters each 1 is .25) then you will get 44.4. To check your answer you would multiply 44.4 and 6.75 and get 299.7. This is not 300 but when you round that .7 to the nearest 10 you will get 300. Hope this helped.

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In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist

WARRIOR [948]

Answer:

(a) The probability that more than 180 will take your free sample is 0.1056.

(b) The probability that fewer than 200 will take your free sample is 0.9997.

(c) The probability that a customer will take a free sample and buy the product is 0.2184.

(d) The probability that between 60 and 80 customers will take the free sample and buy the product is 0.8005.

Step-by-step explanation:

We are given that about 56% of all customers will take free samples. Furthermore, of those who take the free samples, about 39% will buy what they have sampled.

The day you were offering free samples, 303 customers passed by your counter.

Firstly, we will check that it is appropriate to use the normal approximation to the binomial, that is;

Is np > 5 and n(1-p) > 5

In our question, n = sample of customers = 303

p = probability that customers will take free sample = 56%

So, np = $303 \times 0.56$ = 169.68 > 5

n(1-p) = $303 \times (1-0.56)$ = 133.32 > 5

Since, both conditions are satisfied so it is appropriate to use the normal approximation to the binomial.

Now, mean of the normal distribution is given by;

Mean, $\mu$ = $n \times p$ = 169.68

Also, the standard deviation of the normal distribution is given by;

Standard deviation, $\sigma$ = $\sqrt{n \times p \times (1-p)}$

= $\sqrt{303 \times 0.56 \times (1-0.56)}$ = 8.64

Let X = Number of people who will take your free sample

The z score probability distribution for normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

(a) The probability that more than 180 will take your free sample is given by = P(X > 180) = P(X > 180.5) {Using continuity correction}

P(X > 180.5) = P( $\frac{X-\mu}{\sigma}$ > $\frac{180.5-169.68}{8.64}$ ) = P(Z > 1.25) = 1 - P(Z < 1.25)

= 1 - 0.8944 = 0.1056

(b) The probability that fewer than 200 will take your free sample is given by = P(X < 200) = P(X < 199.5) {Using continuity correction}

P(X < 199.5) = P( $\frac{X-\mu}{\sigma}$ < $\frac{199.5-169.68}{8.64}$ ) = P(Z < 3.45) = 0.9997

(c) We are given in the question that of those who take the free samples, about 39% will buy what they have sampled, this means that we have;

P(Buy the product / taken a free sample) = 0.39

So, Probability(customer will take a free sample and buy the product) = P(customer take a free sample) $\times$ P(Buy the product / taken a free sample)

= 0.56 $\times$ 0.39 = 0.2184

(d) Now our mean and standard deviation will get changed because the probability of success now is p = 0.2184 but n is same as 303.

So, Mean, $\mu$ = $n \times p$ = $303 \times 0.2184$ = 66.18

Standard deviation, $\sigma$ = $\sqrt{n \times p \times (1-p)}$

= $\sqrt{303 \times 0.2184 \times (1-0.2184)}$ = 7.192

Now, the probability that between 60 and 80 customers will take the free sample and buy the product is given by = P(60 < X < 80) = P(59.5 < X < 80.5) {Using continuity correction}

P(59.5 < X < 80.5) = P(X < 80.5) - P(X $\leq$ 59.5)

P(X < 80.5) = P( $\frac{X-\mu}{\sigma}$ < $\frac{80.5-66.18}{7.192}$ ) = P(Z < 1.99) = 0.9767

P(X $\leq$ 59.5) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{59.5-66.18}{7.192}$ ) = P(Z $\leq$ -0.93) = 1 - P(Z < 0.93)

= 1 - 0.8238 = 0.1762

Therefore, P(59.5 < X < 80.5) = 0.9767- 0.1762 = 0.8005.

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2 years ago

The ages of Edna, Ellie, and Elsa are consecutive integers. The sum of their ages is 108. What are their ages?

diamong [38]

All I did was guess and check. I started with the 30s because their sum would be nine and there might be a one to carry to make it 10. After many tries (well, not many, I'm just being dramatic) I got 35, 36, and 37. Hope I helped.

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3 years ago

$640,3%,2 years?!?! Plz explain!!

mrs_skeptik [129]

I am pretty sure it is talking about interest and $640 is what you start with, with a 3% interest for 2 years

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2 years ago

a food company delivers its fruit in two types of boxes, large a small. A delivery of 7 large boxes and 9 small boxes has a tota

ivann1987 [24]

Let L and S represent the weights of large and small boxes, respectively. The problem statement gives rise to two equations:
.. 7L +9S = 273
.. 5L +3S = 141

You can solve these equations various ways. Using "elimination", we can multiply the second equation by 3 and subtract the first equation.
.. 3(5L +3S) -(7L +9S) = 3(141) -(273)
.. 8L = 150
.. L = 150/8 = 18.75
Then we can substitute into either equation to find S. Let's use the second one.
.. 5*18.75 +3S = 141
.. S = (141 -93.75)/3 = 15.75

A large box weighs 18.75 kg; a small box weighs 15.75 kg.

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2 years ago

Help pls and quick thanks

storchak [24]

Answer:

I think the Question asks to to take w in left side. so, the answer is below

Step-by-step explanation:

w=sa

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