MATHEMATICS COLLEGE

You can run 10 miles in 80 minutes. How far can you run in 2.5 hours?

Answers

Answer 1

Answer:

Answer:

18.75 miles

Step-by-step explanation:

1 hour = 60 minutes

2.5 Hours = 150 minutes

10 miles _____> 80 minutes

x_______> 150 minutes

x= ( 150 * 10) / 80

= 1500/80

x= 18.75 miles

easy

Answer 2

Answer:

Answer:

Step-by-step explanation:

10 miles- 80 minutes

?- 150 minutes

1500+80x

X= 10.75 miles

Related Questions

What is the sum in simplest form?4 1/2 + 1 3/5
Suppose we have collected times, in minutes, that it takes volunteers to complete a set of pencil and paper mazes. Volunteers are randomly assigned to one of two groups. Group 1 watch a 5 minute video explaining good strategies for completing the mazes. Group 2 watch a 5 minute video of other people successfully completing the mazes, but with no explanation given.Researchers are interested in testing against the null hypothesis that there is no difference in population mean times that it takes people to complete mazes after watching either type of video.There are the summary statistics after collecting data. Note that the sample sizes differ, because some subjects did not follow instructions correctly and so they were not included in the analysis.x1 = 11.98, x2 = 9.15s1 = 8.69, s2 = 7.75n1 = 43, n2 = 52a. Is this an experimental or observational study?
3 3/4 divided by 5/7
You need to purchase juice for a party. The juice is sold for $3 per gallon or $2.50 per quart. You will need 12 oz for each of the 30 students in the class. How much juice will you need
Which expression represents the phrase " the difference of 18 and 12"?

You need to lay communications cable around the perimeter of three offices, each 8' by 10'. How much cable do you need?

Answers

Answer:

108 units

Step-by-step explanation:

We are given that

Length of each office,l=10 units

Breadth of each office=b=8 units

Total number of office=3

We know that

Perimeter of rectangle= $2(l+b)$

Using the formula

Length of cable for 1 office= $2(10+8)=36units$

Length of cable required for 3 offices= $36* 3=108$ units

Hence, the required length of cable=108 units

The Pacheco family has a monthly income of $ 7,300 . They have monthly fixed expenses of $ 3,280 . In July, they had monthly variable expenses of $ 2,620 . They have annual expenses totaling $ 14,000 . Assuming annual expenses are paid off at an equal rate monthly, what was the Pacheco’s cash flow for July?

Answers

Answer:

Positive 233

Step-by-step explanation:

7300-3280-2620-1167(14000/12)=233.

Write an algebraic expression to represent the phrase below.a number n decreased by 5
please help

Answers

Answer

5-n

Step-by-step explanation:

Find an equation of the sphere with center (-3, 2 , 5) and radius 4. What is the intersection of this sphere with the yz-plane.

Answers

Answer:

The equation of the sphere with center (-3, 2 , 5) and radius 4 is $(x+3)^(2) +(y-2)^(2) + (z-5)^(2) = 16$

The intersection of the sphere with the yz- plane gave the equation $(y-2)^(2) + (z-5)^(2) = 7$ which is a 2D- circle with center (0,2,5) and radius $√(7)$ .

Step-by-step explanation:

The equation of a sphere of radius r, with center (a,b,c) is given by

$(x-a)^(2) +(y-b)^(2) + (z-c)^(2) = r^(2)$

where, $x,$ $y,$ and $z$ are the coordinates of the points on the surface of the sphere.

Hence, the equation of the sphere with center, (-3, 2 , 5) and radius 4 becomes

$(x-a)^(2) +(y-b)^(2) + (z-c)^(2) = r^(2)$

$(x-(-3))^(2) +(y-(2))^(2) + (z-(5))^(2) = 4^(2)$

Then,

$(x+3)^(2) +(y-2)^(2) + (z-5)^(2) = 16$

This is the equation of the sphere with center (-3, 2 , 5) and radius 4,

Now, for the intersection of this sphere with the yz- plane,

The $yz -$ plane is where $x = 0$ , then we set $x = 0$

Them the equation $(x+3)^(2) +(y-2)^(2) + (z-5)^(2) = 16$ becomes

$(0+3)^(2) +(y-2)^(2) + (z-5)^(2) = 16$

$(3)^(2) +(y-2)^(2) + (z-5)^(2) = 16\n9 +(y-2)^(2) + (z-5)^(2) = 16\n(y-2)^(2) + (z-5)^(2) = 16 - 9\n(y-2)^(2) + (z-5)^(2) = 7$

This equation is the equation of a 2D- circle with center (0,2,5) and radius $√(7)$

This is the part of the sphere that intersects with the yz-plane.

6. Jefferson has $30,000 property damage insurance and comprehensive/collision insurance with adeductible of $1,000. He crashes into a house causing $40,000 damage to the house. Jefferson's car needs
$4,500 worth of repairs? How much will Jefferson have to pay total to fix the house and his car?

Answers

Answer: Since Jefferson has a $1,000 deductible on his comprehensive/collision insurance, he will have to pay the first $1,000 of his car repairs himself. The insurance will cover the remaining amount, which is $4,500 - $1,000 = $3,500.

For the damage to the house, Jefferson's property damage insurance will cover up to $30,000. Since the damage to the house is $40,000, Jefferson will have to pay the difference out of his own pocket. The amount he will have to pay is $40,000 - $30,000 = $10,000.

Therefore, the total amount Jefferson will have to pay to fix the house and his car is $1,000 + $3,500 + $10,000 = $14,500.

Step-by-step explanation:

Solbe for x : x-1/x+3x+3=0

Answers

Answer:

x=1/4 or x=-1

Step-by-step explanation:

x−

+3x+3=0

4x2+3x−1

Step 1: Multiply both sides by x.

4x2+3x−1=0

(4x−1)(x+1)=0(Factor left side of equation)

4x−1=0 or x+1=0(Set factors equal to 0)

or x=−1

Check answers. (Plug them in to make sure they work.)

(Works in original equation)

x=−1(Works in original equation)

Answer:

or x=−1

Other Questions

Q 3.28: To create a confidence interval from a bootstrap distribution using percentiles, we keep the middle values and chop off a certain percent from each tail. Indicate what percent of values must be chopped off from each tail for a 97% confidence level. A : We keep the middle 97% of values by chopping off 1.5% from each tail. B : We keep the middle 1.5% of values by chopping off 97% from each tail. C : We keep the middle 3% of values by chopping off 97% from each tail. D : We keep the middle 3% of values by chopping off 1.5% from each tail. E : We keep the middle 97% of values by chopping off 3% from each tail.

The state department of transportation is coordinating road crews to fix potholes after a particularly snowy winter. Initial estimates gave an average of 7.8 potholes per mile of highway. Find the probability that there are 11 or more potholes in a randomly selected one-mile stretch of highway. Use Excel to find the probability.

PLEASE HELP IF YOU KNOW THIS!!!!!

Find y from the picture plz