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2 years ago

13

What is the volume of the pyramid? A 17cm^3, 56cm^3, 84cm^3 or 168cm^3

1 answer:

Alborosie2 years ago

7 0

Answer:

56cm2

Step-by-step explanation:

1÷3×4×6×7

=56cm2

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The principal of a middle school claims that test scores of the seventh-graders at his school vary less than the test scores of

AveGali [126]

Answer:

There is sufficient evidence to support the claim that the standard deviation is less than 14.7.

Step-by-step explanation:

From the question we are told that that the null hypothesis is

$H_o : \sigma < 14.7$

Now given from the question that this null hypothesis was rejected, it mean, in non technical term that there is sufficient evidence to support the claim that the standard deviation is less than 14.7.

3 0

3 years ago

The average value of a function f over the interval [−2,3] is −6 , and the average value of f over the interval [3,5] is 20. Wha

Xelga [282]

Answer:

The average value of $f$ over the interval $[-2,5]$ is $\frac{10}{7}$ .

Step-by-step explanation:

Let suppose that function $f$ is continuous and integrable in the given intervals, by integral definition of average we have that:

$\frac{1}{3-(-2)} \int\limits^{3}_{-2} {f(x)} \, dx = -6$ (1)

$\frac{1}{5-3} \int\limits^{5}_{3} {f(x)} \, dx = 20$ (2)

By Fundamental Theorems of Calculus we expand both expressions:

$\frac{F(3)-F(-2)}{3-(-2)} = -6$

$F(3) - F(-2) = -30$ (1b)

$\frac{F(5)-F(3)}{5-3} = 20$

$F(5) - F(3) = 40$ (2b)

We obtain the average value of $f$ over the interval $[-2, 5]$ by algebraic handling:

$F(5) - F(3) +[F(3)-F(-2)] = 40 + (-30)$

$F(5) - F(-2) = 10$

$\frac{F(5)-F(-2)}{5-(-2)} = \frac{10}{5-(-2)}$

$\bar f = \frac{10}{7}$

The average value of $f$ over the interval $[-2,5]$ is $\frac{10}{7}$ .

4 0

2 years ago

Here goes another question lol oops

Murrr4er [49]

Student 1 and 3.
$- (5r - 3s + 1) = - 5r + 3s - 1$
Move the negative into the brackets,
positive becomes negative while two negatives make a positive.

3 0

3 years ago

Read 2 more answers

Devon made a box with length x+1, width x+3, and height x-3. What is the volume of Devon's box as a function of x?

erica [24]

Answer:

(a)

$V=x^3+x^2-9x-9$

(b)

$x=10$

(c)

$x=\frac{7}{2}$

Step-by-step explanation:

we are given

length is x+1

$L=x+1$

width is x+3

$W=x+3$

height is x-3

$H=x-3$

(a)

we can use volume formula

$V=L\times W\times H$

now, we can plug it

$V=(x+1)\times (x+3)\times (x-3)$

now, we can simplify it

$V=x^3+x^2-9x-9$

(b)

we are given volume =1001

so, we can set V=1001

and then we can solve for x

$V=x^3+x^2-9x-9=1001$

now, we can factor it

$\left(x-10\right)\left(x^2+11x+101\right)=0$

$x-10=0$

$x=10$

(b)

we are given volume

so, we can set

$V=14\frac{5}{8}=\frac{14\times 8+5}{8}$

$V=\frac{117}{8}$

and then we can solve for x

$V=x^3+x^2-9x-9=\frac{117}{8}$

now, we can factor it

$8x^3+8x^2-72x-189=0$

$\left(2x-7\right)\left(4x^2+18x+27\right)=0$

$2x-7=0$

$x=\frac{7}{2}$

8 0

3 years ago

Please help solve <br><br> -2x - 4 - 7 = - 3x - 15

valentina_108 [34]

Answer:

x = -4

Step-by-step explanation:

-2x-11=-3x-15

x - 11 = -15

x = -4

8 0

2 years ago