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madreJ [45]
2 years ago
5

If a car travels at an average speed of x miles per hour, how far would the car travel in 90 minutes?

Mathematics
1 answer:
Anastasy [175]2 years ago
5 0

Answer:

1.5(x)

Step-by-step explanation:

find out how many hours are in 90 minutes

1.5

x miles

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What is the volume of the square pyramid with base edges 32 mm and slant height 34 mm?
NeX [460]

Answer:

The answer to your question is 10240 mm³

Step-by-step explanation:

Data

length of the base = 32 mm

length of the height = 34 mm

Formula

Volume of a pyramid = 1/3 x Area of the base x length of the height

Process

1.- Calculate the area of the base

Area = side x side

        = 32 x 32

        = 1024 mm²

2.- Find the height of the pyramid using the Pythagorean theorem

height² = 34² - 16²

height² = 1156 - 256

height² = 900

height = 30

3.- Calculate the volume of the pyramid

Volume = 1/3Area x height

             = 1/3(1024 x 30)

             = (30720)/3 mm³

             = 10240 mm³

5 0
3 years ago
A table titled Number of Dogs Adopted has entries 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9. The total is 13
saw5 [17]

Answer:

1) Mean = 6.28

2) Median = 7

3) Mode = 8

4) Range = 6

Step-by-step explanation:

We need to find Mean, Median, Mode and Range of data given in tables.

1) Mean

The formula used to calculate mean is: Mean=\frac{Sum\:of\:all\:data\:points}{Number\:of\:data\:points}

The data given is: 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9

Sum of all data points = 132

Number of data points = 21

Finding mean:

Mean=\frac{Sum\:of\:all\:data\:points}{Number\:of\:data\:points}\\Mean=\frac{132}{21}\\Mean=6.28

So, Mean = 6.28

2) Median

The formula used to calculate median is: Median=\frac{n+1}{2}\:th\: term because we have n = odd

Number of data points n = 21

Putting values and finding the position of median term

Median=\frac{n+1}{2}\:th\: term\\Median=\frac{21+1}{2}\:th\: term\\Median=\frac{22}{2}\:th\: term\\Median=11\:th\:term

So, in the given data : 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9

The 11 th term is 7

So, Median = 7

3) Mode

The mode is the most repetitive value of data set.

In the given data set: 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9

The most repetitive value is 8

So, Mode = 8

4) Range

The range can be calculated using formula: Range=Maximum\:Value-Minimum\:Value

In the given data set: 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9

Maximum value = 9

Minimum value = 3

So, the range is:

Range=Maximum\:Value-Minimum\:Value\\Range=9-3\\Range =6

So, Range = 6

3 0
3 years ago
Read 2 more answers
CREATE AN EXPRESSION & SOLVE.
Harrizon [31]

Answer:

24/3-4

Answer is 12

Step-by-step explanation:

To find the answer, you have to work backwards. You also have to use PEMDAS. Since she multiplied her answer to get 24, you have to work backwards and divide by 3 first. You will get the number 8 after that because 24/3 is 8. Since she also subtracted 4, you have to work backwards and add 4. That gives you an original number of 12.

3 0
2 years ago
Read 2 more answers
49.5 is what percent of 33?
Ivan
\begin{array}{ccc}33&-&100\%\\\\49.5&-&p\%\end{array}\\\\cross\ multiply\\\\33\cdot p=49.5\cdot100\ \ \ \ /:33\\\\p=\frac{4950}{33}\\\\p=150\\\\Answer:150\%.
5 0
3 years ago
Read 2 more answers
Someone please help Me! Solve each system of equation. I need help I don’t understand and this is due tomorrow. I’ll make brainl
mestny [16]

Answer:

  (a, b, c) = (-2, -3, 5)

Step-by-step explanation:

The idea is to find values of a, b, and c that satisfy all three equations. Any of the methods you learned for systems of 2 equations in 2 variables will work with 3 equations in 3 variables. Often, folks find the "elimination" method to be about the easiest to do by hand.

Here, we notice that the coefficients in general are not nice multiples of each other. However, the coefficient of "a" in the second equation is 1, so we can use that to eliminate "a" from the other equations.

__

Add 3 times the second equation to the first.

  3(a +3b -4c) +(-3a -4b +2c) = 3(-31) +(28)

  3a +9b -12c -3a -4b +2c = -93 +28 . . . . eliminate parentheses

  5b -10c = -65 . . . . . . . . . . . . . . . . . . . . . . .collect terms

  b - 2c = -13 . . . . . . . divide by 5 (because we can) "4th equation"

Subtract 2 times the second equation from the third.

  (2a +3c) -2(a +3b -4c) = (11) -2(-31)

  2a +3c -2a -6b +8c = 11 +62

  -6b +11c = 73 . . . . . . "5th equation"

Now, we have two equations in b and c that we can solve in any of the ways we know for 2-variable equations. Once again, it looks convenient to use the first of these (4th equation) to eliminate the b variable. (That is why we made its coefficient be 1 instead of leaving it as 5.)

Add 6 times the 4th equation to the 5th equation:

  6(b -2c) + (-6b +11c) = 6(-13) +(73)

  6b -12c -6b +11c = -78 +73 . . . . . . . eliminate parentheses

  -c = -5 . . . . . . . . . . . . . . . . . . . . . . . . collect terms

  c = 5 . . . . . multiply by -1

We can now work backwards to find the other variable values. Substituting into the 4th equation, we have ...

  b -2(5) = -13 . . . .substitute for c

  b = -3 . . . . . . . . . add 10

And the values for b and c can be substituted into the 2nd equation.

  a + 3(-3) -4(5) = -31

  a -9 -20 = -31 . . . . . eliminate parentheses

  a = -2 . . . . . . . . . . . . add 29

__

The solution to this set of equations is (a, b, c) = (-2, -3, 5).

_____

<em>Comment on steps</em>

At each stage, we made choices calculated to simplify the process. By using equations that had a variable coefficient of 1, we avoided messy fractions or multiplying by more numbers than necessary when we used the elimination process. That is, the procedure is guided by the idea of <em>elimination</em>, but the specific steps are <em>ad hoc</em>. Using <em>these same specific steps</em> on different equations will likely be useless.

_____

<em>Alternate solution methods</em>

The coefficients of these equations can be put into the form called an "augmented matrix" as follows:

\left[\begin{array}{ccc|c}-3&-4&2&28\\1&3&-4&-31\\2&0&3&11\end{array}\right]

Many graphing and/or scientific calculators are able to solve equations written in this form. The function used is the one that puts this matrix into "reduced row-echelon form". The result will look like ...

\left[\begin{array}{ccc|c}1&0&0&-2\\0&1&0&-3\\0&0&1&5\end{array}\right]

where the rightmost column is the solution for the variables in the same order they appear in the equations. The vertical line in the body of the matrix may or may not be present in a calculator view. The square matrix to the left of the vertical bar is an identity matrix (1 in each diagonal element) when there is exactly one solution.

6 0
3 years ago
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