MATHEMATICS HIGH SCHOOL

A van travels 220 miles on 10 gallons of gas. Find how many gallons the van needs to travel 550 miles

Answers

Answer 1

Answer:

Answer: The required number of gallons to travel 550 miles is 25.

Step-by-step explanation: Given that a van travels 220 miles on 10 gallons of gas.

We are to find the number of gallons that the van needs to travel 550 miles.

We will be using the UNITARY method to solve the given problem.

We have

Number of gallons required to travel 220 miles = 10.

So, number of gallons required to travel 1 mile $=(10)/(220).$

Therefore, the number of gallons required to travel 550 gallons will be

$(10)/(220)* 550=(550)/(22)=25.$

Thus, the required number of gallons to travel 550 miles is 25.

Answer 2

Answer: Use property of proportion:

220 mi ----------> 10 gal
550 mi ----------> x

$x=(10*550)/(220)=25 \ gallons$

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which of the following expressions is equivalent to 20 - 4/5 x > (the > has a line under it) 16?

Answers

$20-(4)/(5)x\geq16\ \ \ \ |both\ sides\ -20\n\n-(4)/(5)x\geq-4\ \ \ \ |both\ sides\ :(-(5)/(4)) < 0\ then\ change\$

(x - 5)2 = 25Solution steps:
Add 5 to both sides
Multiply both sides by 5
Square both sides
Take the square root of both sides

Answers

Here is your answer

x = 17.5

Make as brainliest plzz....

Hope this answer is helpful....

The average high temperature in Anchorage, Alaska, in January is 21°F with a standard deviation of 10°. The average high temperature in Honolulu in January is 80°F with a standard deviation of 8°. In which location would it be more unusual to have a day in January with a high of 57°F?

Answers

Answer:

In Honolulu it be more unusual to have a day in January with a high of 57°F

Step-by-step explanation:

Given,

The average high temperature in Anchorage, Alaska,

$\mu_1$ = 21° F,

Having standard deviation,

$\sigma_1$ = 10°,

Thus, the z-score for the high temperature of 57° F in alaska,

$z_1=(57-\mu_1)/(\sigma_1)$

$=(57-21)/(10)$

$=3.6$

By the z-score table the probability of the high temperature of 57°F in Alaska = 0.99984 = 99.984 %

Now, the average high temperature in Honolulu is,

$\mu_2=80^(\circ)$

Having standard deviation,

$\sigma_2=8^(\circ)$

Thus, the z-score for the high temperature of 57° F in Honolulu,

$z_2=(57-\mu_2)/(\sigma_2)$

$=(57-80)/(8)$

$=-2.875$

By the z-score table the probability of the high temperature of 57° in Honolulu = 0.00205 = 0.205 %

Since, a probability is unusual if it is less than 5 %,

Hence, In Honolulu it be more unusual to have a day in January with a high of 57°F.

It would be more odd in Honolulu.

What ways can you collect 82 cents ? Using 13 coins .

Answers

saving up or just ask your mom for miney

(6x^5+21x^4+33x^3+0x^3+18x+12)÷(6x+3)

Answers

so first factor out the big equation on top
very hard
we can cancel out the 0x terms since 0x=0

so we notice that 3 is a factor of both equations
(6x+3)=(3)(2x+1)

and
(6x^5+21x^4+33x^3+18x+12)=(3)(2x+1)(x^4+3x^3+4x^2-2x+4
so we notice that there is a (3)(2x+1) in bot h so divide and cross them out

the answer is (x^4+3x^3+4x^2-2x+4)

$(2a + 4b)/(2b) = 3.9 \n (a)/(b) = what$
with full explanation please

Answers

Answer: 1.9

Work Shown:

$(2a+4b)/(2b) = 3.9\n\n(2a)/(2b)+(4b)/(2b) = 3.9\n\n(a)/(b)+2 = 3.9\n\n(a)/(b) = 3.9-2\n\n(a)/(b) = 1.9\n\n$

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