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3 years ago

When x = 3 and y = 5, by how much does the value of 3x2 – 2y exceed the value of 2x2 – 3y ?

2 answers:

8 0

The answer goes up by 14 because the first problem equals 8 and the second problem equals --3 therefor the difference is 14

fgiga [73]3 years ago

4 0

$3\cdot3^2-2\cdot5=27-10=17\\
2\cdot3^2-3\cdot5=18-15=3\\\\
17-3=14$

By 14.

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21 = -3c + 7 -5c -2 i need help with this multi step equation question

belka [17]

Answer:

c=-2

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

I need to understand question 14 and 15 and the last one cut off for tens in one

sp2606 [1]

Question # 14

Given the numbers

10 11 12 13 14 15 16 17 18 19 20

Let 'x' be the number

The condition breakdown:

I am less than 20.

So the number 'x' must be less than 20 i.e. x < 20

I am more than 13.

So the number 'x' must be greater than 13 i.e. x > 13

I am less than 17.

So the number 'x' must be less than 17 i.e. x < 17

Finally:

I am 4 more than 12

i.e. 12+4 = 16

Thus, the number is x = 16

Question # 15

Part a)

Given the numbers

10 11 12 13 14 15 16 17 18 19 20

Let 'x' be the number

The condition breakdown:

I am more than 10.

x > 10

I am less than 20.

x < 20

I am more than 12.

x > 12

I am less than 15.

x < 15

As the numbers left after all the conditions are fulfilled are 13 and 14.

But the last condition is, of the numbers that left, the number is greater than all the remaining numbers.

So, from the remaining number 13 and 14;

14 > 13

Thus, the number x = 14

Part b)

Drawing the number 14 in the place value:

Chart

Tens Ones

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6 0

3 years ago

What goes in the blank? A always B sometimes C never D rarely

o-na [289]

I think it might be (b)

6 0

3 years ago

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A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and

pav-90 [236]

Answer:

ans=13.59%

Step-by-step explanation:

The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean $\mu$ and standard deviation $\sigma$ , we have these following probabilities

$Pr(\mu - \sigma \leq X \leq \mu + \sigma) = 0.6827$

$Pr(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545$

$Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973$

In our problem, we have that:

The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months

So $\mu = 53, \sigma = 11$

So:

$Pr(53-11 \leq X \leq 53+11) = 0.6827$

$Pr(53 - 22 \leq X \leq 53 + 22) = 0.9545$

$Pr(53 - 33 \leq X \leq 53 + 33) = 0.9973$

-----------

$Pr(42 \leq X \leq 64) = 0.6827$

$Pr(31 \leq X \leq 75) = 0.9545$

$Pr(20 \leq X \leq 86) = 0.9973$

-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have $Pr(31 \leq X \leq 75) = 0.9545 = 0.9545$ subtracted by $Pr(42 \leq X \leq 64) = 0.6827$ is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.

To find just the percentage above the mean, we divide this value by 2

So:

$P = {0.9545 - 0.6827}{2} = 0.1359$

The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.

4 0

4 years ago

Find the distance between the points L(7,-1) and M(-2, 4).

zysi [14]

Answer:

The answer is

$\sqrt{106} \: \: \: or \: \: \: 10.3 \: \: \: \: units$

Step-by-step explanation:

The distance between two points can be found by using the formula

$d = \sqrt{ ({x1 - x2})^{2} + ({y1 - y2})^{2} } \\$

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

L(7,-1) and M(-2, 4)

The distance between them is

$|LM| = \sqrt{ ({7 + 2})^{2} + ( { - 1 - 4})^{2} } \\ = \sqrt{ {9}^{2} + ({ - 5})^{2} } \\ = \sqrt{81 + 25} \\ = \sqrt{106} \: \: \: \: \: \: \: \\ = 10.29563014...$

We have the final answer as

$\sqrt{106} \: \: \: or \: \: \: 10.3 \: \: \: \: units$

Hope this helps you

4 0

3 years ago