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3 years ago

A car travels 200 miles using 7 gallons of gas. At that rate, how far can the car travel using 35 gallons of gas?

Mathematics

1 answer:

Eduardwww [97]3 years ago

7 0

Answer:

7 x 5 = 35

They are asking how far the car can go if you have 5 times the amount of gas.

200 x 5 = 1,000 miles

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3 years ago

Operations with rational expressions

Scorpion4ik [409]

$$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}=-\frac{w+7}{9}$

Solution:

Given expression:

$$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}$

To solve the given expression:

First simplify: $\frac{50-2 w^{2}}{3 w^{2}+9 w-30}$

$$\frac{50-2 w^{2}}{3 w^{2}+9 w-30}=-\frac{2(w+5)(w-5)}{3(w-2)(w+5)}$

Cancel the common factor (w + 5).

$$=-\frac{2(w-5)}{3(w-2)}$

Now substitute this in the given expression.

$$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}=-\frac{2(w-5)}{3(w-2)} \cdot \frac{w^{2}+5 w-14}{6 w-30}$

Multiply the fractions $\frac{a}{b} \cdot \frac{c}{d}=\frac{a \cdot c}{b \cdot d}$

$$=-\frac{2(w-5)\left(w^{2}+5 w-14\right)}{3(w-2)(6 w-30)}$

Factor the denominator $3(w-2)(6 w-30) =18(w-2)(w-5)$

$$=-\frac{2(w-5)\left(w^{2}+5 w-14\right)}{18(w-2)(w-5)}$

Cancel the common factor 2(w – 5).

$$=-\frac{w^{2}+5 w-14}{9(w-2)}$

Factor the numerator $w^{2}+5 w-14=(w-2)(w+7)$

$$=-\frac{(w-2)(w+7)}{9(w-2)}$

Cancel the common factor (w – 2).

$$=-\frac{w+7}{9}$

$$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}=-\frac{w+7}{9}$

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3 years ago

What is the statistical decision regarding the differences between the observed and expected frequencies if the critical value o

mel-nik [20]

Answer:

Fail to reject.

Step-by-step explanation:

For a goodness of fit test the decision rule is:

If the calculated value of the test statistic is more than the right tailed critical value of chi-square, then the null hypothesis will be rejected.

The hypothesis for a goodness of fit test is:

<em>H</em>₀: There is no differences between the observed and expected frequencies.

<em>Hₐ</em>: There is a significant differences between the observed and expected frequencies.

The computed value of the test statistic is, 6.079.

The critical value of chi-square is 9.488.

Calculated test statistic < Critical value

6.079 < 9.488

Thus, the null hypothesis was failed to rejected.

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3 years ago

What is the slope of this line?

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Answer:

-1 is the slope

Step-by-step explanation:

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3 years ago

What is 2.48 as a whole number

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ANSWER 2.48 as a whole number would be 2 because you would round to the nearest whole number. GIVE BRAINLIEST PLEASE

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3 years ago

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