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3 years ago

8

2. Evaluate the expression by substituting the given value(s):

1 answer:

lawyer [7]3 years ago

7 0

Answer:

C

is you answers because i did the math so u dont have to ;)

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Please help ASAP! Due in a hour! :) answer both and I will thank and mark as brainliest

lorasvet [3.4K]

Answer:

22: 45 and 126

Step-by-step explanation:

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3 years ago

I’m confused on this one

Yakvenalex [24]

It would be 10 inches. The length of WZ would usually be stated as the diameter in these cases. As you can see in the picture, 5 inches is our radius and the diameter equals the radius times two (2r or the radius doubled) to give the diameter. 5 x 2 = 10, and 10 would be the answer.

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4 years ago

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topjm [15]

I believe the answer is 28.27, or 28.3

5 0

3 years ago

Calculate f(x) for the given domain values.

Maksim231197 [3]

Answer:

See explanation

Step-by-step explanation:

1. The given function is

$f(x) = - 3x$

The domain values are: x=0, 2, -1, 4, -2

When x=0

$f(0) = - 3 \times 0 = 0$

When x=2,

$f(x) = - 3 \times 2 = - 6$

When x=-1

$f(x) = - 3 \times - 1 = 3$

When x=4

$f(4) = - 3 \times 4 = - 12$

When x=-2

$f( - 2) = - 3 \times - 2 = 6$

2. The given function is

$f(x) = \frac{1}{3} x$

When x=3,

$f(3) = \frac{1}{3} \times 3 = 1$

Similarly,

$f( - 3) = \frac{1}{3} \times - 3 = - 1$

$f(300) = \frac{1}{3} \times 300 = 100$

$f( - 180) = \frac{1}{3} \times - 180 = - 60$

$f(99) = \frac{1}{3} \times 99 = 33$

8 0

3 years ago

Consider the following ordered data. 6 9 9 10 11 11 12 13 14 (a) Find the low, Q1, median, Q3, and high. low Q1 median Q3 high (

IrinaVladis [17]

Answer:

Low Q1 Median Q3 High

6 9 11 12.5 14

The interquartile range = 3.5

Step-by-step explanation:

Given that:

Consider the following ordered data. 6 9 9 10 11 11 12 13 14

From the above dataset, the highest value = 14 and the lowest value = 6

The median is the middle number = 11

For Q1, i.e the median of the lower half

we have the ordered data = 6, 9, 9, 10

here , we have to values as the middle number , n order to determine the median, the mean will be the mean average of the two middle numbers.

i.e

median = $\dfrac{9+9}{2}$

median = $\dfrac{18}{2}$

median = 9

Q3, i.e median of the upper half

we have the ordered data = 11 12 13 14

The same use case is applicable here.

Median = $\dfrac{12+13}{2}$

Median = $\dfrac{25}{2}$

Median = 12.5

Low Q1 Median Q3 High

6 9 11 12.5 14

The interquartile range = Q3 - Q1

The interquartile range = 12.5 - 9

The interquartile range = 3.5

7 0

3 years ago