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erastovalidia [21]
2 years ago
7

Silvergrove middle school spent $60 to purchase 4 work books. How many these workbooks can they buy for $90? Solve using unit ra

tes.
Mathematics
1 answer:
cluponka [151]2 years ago
6 0
They can buy 6 workbooks with $90. To find the unit rate we simply divide the cost ($60) by the number of books (4). That gives us $15, the cost of 1 workbook. So now to find out how many workbooks can be purchased with $90, we divide 90 by 15. That’s how I got 6 workbooks. Hope this helps.
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Susan paid $0.78 per liter for gas while driving across Canada. Find the cost per gallon to the nearest cent.
denis-greek [22]

Answer:

<u>The cost of gas per gallon in Canada is $ 2.95 (Rounding to the nearest cent)</u>

Step-by-step explanation:

Let's solve this question, this way:

1. Let's convert liters to US gallons:

1 US gallon = 3.785412 liters

1 liter = 0. 26417204 gallon

2. Let's calculate the cost per gallon:

1 liter of gas = $ 0.78

3.785412 liters = 0.78 * 3.785412

1 gallon = $ 2.95 (Rounding to the nearest cent)

<u>The cost of gas per gallon in Canada is $ 2.95 (Rounding to the nearest cent)</u>

7 0
3 years ago
Read 2 more answers
Aiden calculated the area of Triangle QRS finding half the product of 1.5 and 1.1. What error did Aiden make?
Gnom [1K]
Aiden should have calculated 
area = (1.1 * 5.4) / 2
because triangle area = half the base times the altitude.

6 0
3 years ago
74% of 19 year old males are atleast 172 pounds , what percent of the males are not atleast 172 pounds?
swat32

Answer:

<em>The percentage of males are not at least 172 pounds</em>

P(X⁻ ≥ 172) = 0.26

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given that 74% of 19 -year -old males are at least 172 pounds

Let 'X' be a random variable in a binomial distribution

P( X≥172) = 74% = 0.74

<em>we have to find that the percentage of males are not at least 172 pounds</em>

<u><em>Step(ii):-</em></u>

<em>The probability of males are not at least 172 pounds</em>

P(X⁻≥172) = 1- P( X≥172)

                 = 1- 0.74

<em>                  = 0.26</em>

<u><em>Final answer:-</em></u>

<em>The percentage of males are not at least 172 pounds</em>

P(X⁻ ≥ 172) = 0.26

<u><em></em></u>

7 0
2 years ago
Of the total population of American households, including older Americans and perhaps some not so old, 17.3% receive retirement
Alex Ar [27]

Answer:

47.54% probability that more than 20 households but fewer than 35 households receive a retirement income

Step-by-step explanation:

We use the binomial aproxiation to the normal to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

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.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.173, n = 120. So

\mu = E(X) = np = 120*0.173 = 20.76

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{120*0.173*0.827} = 4.14

In a random sample of 120 households, what is the probability that more than 20 households but fewer than 35 households receive a retirement income?

We are working with discrete values, so this is the pvalue of Z when X = 35-1 = 34 subtracted by the pvalue of Z when X = 20 + 1 = 21.

X = 34

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

Z = \frac{34 - 20.76}{4.14}

Z = 3.2

Z = 3.2 has a pvalue of 0.9993

X = 21

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

Z = \frac{21 - 20.76}{4.14}

Z = 0.06

Z = 0.06 has a pvalue of 0.5239

0.9993 - 0.5239 = 0.4754

47.54% probability that more than 20 households but fewer than 35 households receive a retirement income

6 0
3 years ago
PLEASSEEEEEE HELPPPPPP! WILL GIVE BRAINLIEST.
4vir4ik [10]

First replace x and y by 3 and 4.

(3*(x)+1)/(4*(y)^2)

((3*(3))+1)/(4*((4)^2)) = (9+1)/(4*16) = 10/64 = 5/32.

5/32 is your answer.

Please mark me as brainliest!

3 0
3 years ago
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