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3 years ago

How many miles does it travel in 1 hour?

Mathematics

2 answers:

podryga [215]3 years ago

8 0

It would be 2.1 hours or about 2 hours.

It says the train travels about 2118 miles per 8 miles. So, divide 2118 by 8, which gives you 264.75 (when you see the word PER it usually indicates to divide). That gives you part A.

Then, you divide 560, (which is t<span>he distance between Columbus, Ohio, and New York) by 264.75 to see how many hours it took, which gives you 2.1152030217...., which simplifies to 2.1. Which gives you the answer for B.

So,
A is 560 and
B is 2.1 or 2

</span>

Naya [18.7K]3 years ago

3 0

If you sent up a proportion, 8/144
and simplify it to the unit rate, you get 1/268
and then set up the other fraction x/560.
divide 560 by 268 you get the answer of 2.089... hours

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3 years ago

Dr. Potter provides vaccinations against polio and measles. Each polio vaccination consists of 6 doses, and each measles vaccina

Lynna [10]

Answer:

The number of measles vaccines that Dr. Potter give than polio vaccines is 30

Step-by-step explanation:

The parameters given are;

The number of doses given in a polio vaccine = 6 doses

The number of doses given in a measles vaccine = 3 doses

The number of vaccinations given by Dr. Potter last year = 60 vaccinations

The number of doses given in the 60 vaccinations = 225 doses

Let the number of polio vaccine given last year by Dr. Potter = x

Let the number of measles vaccine given last year by Dr. Potter = y

Therefore, we have;

6 × x + 3 × y = 225.......................(1)

x + y = 60.......................................(2)

From equation (2), we have;

x = 60 - y

Substituting the derived value for x in equation (1), we get;

6 × x + 3 × y = 225

6 × (60 - y) + 3 × y = 225

360 - 6·y + 3·y = 225

360 - 225 = 6·y - 3·y

135 = 3·y

y = 45

x = 60 - y = 60 - 45 = 15

Therefore;

The number of polio vaccine given last year by Dr. Potter = 15

The number of measles vaccine given last year by Dr. Potter = 45

The number of measles vaccines that Dr. Potter give than polio vaccines = 45 - 15 = 30 vaccines.

The number of measles vaccines that Dr. Potter give than polio vaccines = 30 vaccines.

7 0

3 years ago

The distribution of weights for newborn babies is approximately normally distributed with a mean of 7.4 pounds and a standard de

blsea [12.9K]

Answer:

1. 15.87%

2. 6 pounds and 8.8 pounds.

3. 2.28%

4. 50% of newborn babies weigh more than 7.4 pounds.

5. 84%

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 7.4 pounds

Standard Deviation, σ = 0.7 pounds

We are given that the distribution of weights for newborn babies is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

1.Percent of newborn babies weigh more than 8.1 pounds

P(x > 8.1)

$P( x > 8.1) = P( z > \displaystyle\frac{8.1 - 7.4}{0.7}) = P(z > 1)$

$= 1 - P(z \leq 1)$

Calculation the value from standard normal z table, we have,

$P(x > 8.1) = 1 - 0.8413 = 0.1587 = 15.87\%$

15.87% of newborn babies weigh more than 8.1 pounds.

2.The middle 95% of newborn babies weight

Empirical Formula:

Almost all the data lies within three standard deviation from the mean for a normally distributed data.
About 68% of data lies within one standard deviation from the mean.
About 95% of data lies within two standard deviations of the mean.
About 99.7% of data lies within three standard deviation of the mean.

Thus, from empirical formula 95% of newborn babies will lie between

$\mu-2\sigma= 7.4-2(0.7) = 6\\\mu+2\sigma= 7.4+2(0.7)=8.8$

95% of newborn babies will lie between 6 pounds and 8.8 pounds.

3. Percent of newborn babies weigh less than 6 pounds

P(x < 6)

$P( x < 6) = P( z > \displaystyle\frac{6 - 7.4}{0.7}) = P(z < -2)$

Calculation the value from standard normal z table, we have,

$P(x < 6) =0.0228 = 2.28\%$

2.28% of newborn babies weigh less than 6 pounds.

4. 50% of newborn babies weigh more than pounds.

The normal distribution is symmetrical about mean. That is the mean value divide the data in exactly two parts.

Thus, approximately 50% of newborn babies weigh more than 7.4 pounds.

5. Percent of newborn babies weigh between 6.7 and 9.5 pounds

$P(6.7 \leq x \leq 9.5)\\\\ = P(\displaystyle\frac{6.7 - 7.4}{0.7} \leq z \leq \displaystyle\frac{9.5-7.4}{0.7})\\\\ = P(-1 \leq z \leq 3)\\\\= P(z \leq 3) - P(z < -1)\\= 0.9987 -0.1587= 0.84 = 84\%$