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3 years ago

1 point

Mathematics

1 answer:

denis23 [38]3 years ago

8 0

Answer:

128cm

Step-by-step explanation:

since there are 4 triangles that have an area of 8*4 each, you can multiply 8*4 together to get 32. Then you would multiply 32 by 4 to get the entire area of the whole figure.

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What does the equation x^2+y^2=14 represent?

sdas [7]

Your answer is b. circle

8 0

3 years ago

Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval?

liq [111]

Answer:

7/3

Step-by-step explanation:

f(x) = 3x^2 + 2x + 1, [0, 2]

f(0)=1

f(2)=17

f'(c)=f(2)-f(0)/2-1

f'(c)=17-1/1=16

f'(c)= 6x+2

6x+2=16

6x=16-2

6x=14

x=14/6

x=7/3

4 0

3 years ago

Read 2 more answers

Lori rented a car for one day. The daily rental rate was $12.50 plus 25 cents per mile. The total cost of her rental was $46. Ho

Leona [35]

Answer:

The number of miles Lori drive be 134 miles .

Step-by-step explanation:

Let us assume that the number of miles Lori be x.

As given

Lori rented a car for one day.

The daily rental rate was $12.50 plus 25 cents per mile.

As 1 dollar = 100 cent

$1\ cent = \frac{1}{100}\ dollar$

$25\ cent = \frac{25}{100}\ dollar$

$25\ cent = 0.25\ dollar$

As given

The total cost of her rental was $46.

Than the equation becomes

46 = 12.50 + 0.25x

46 - 12.50 = 0.25x

33.5 = 0.25x

$x = \frac{33.5}{0.25}$

x = 134 miles

Therefore the number of miles Lori drive be 134 miles .

6 0

2 years ago

In a random sample of 18 families, the average weekly food expense was $95.60 with a sample standard deviation of $22.50. Determ

shepuryov [24]

Answer:

t-distribution should be used to construct a confidence interval.

Step-by-step explanation:

We are given that a random sample of 18 families, the average weekly food expense was $95.60 with a sample standard deviation of $22.50.

We have to determine whether a normal distribution (Z values) or a t- distribution should be used or whether neither of these can be used to construct a confidence interval.

Since in this question we are provided with;

Sample average weekly food expense, $\bar X$ = $95.60

Sample standard deviation, s = $22.50

Sample of families, n = 18

The distribution that we will use here to construct a confidence interval will be t-distribution because in the question we don't know anything about population standard deviation $(\sigma)$ .