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4 years ago

What is the base of a triangle whose height is 10 and whose area is 160

1 answer:

5 0

B x h = triangle area
triangle area= 160
height= 10
b x 10 (divide it by) 2= b
160 x 2 (divide it by) 10= b
32= b
the height of the this triangle is 32

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Please help i will thank and rate 5 star!

stich3 [128]

The sign of the product of that equation would be negative.

5 0

3 years ago

A+44=B A=1x+76 B=-6x+134 What is the value of A?

Rashid [163]

Hi there! A = 78

A + 44 = B
We can plug in the values of A and B into the equation. We then get an equation with only one variable (x), which we can solve.

1x + 76 + 44 = - 6x + 134
Collect terms.

1x + 120 = - 6x + 134
Add 6x to both sides.

7x + 120 = 134
Subtract 120 from both sides

7x = 14
Divide both sides by 7

x = 14 / 7 = 2

A = 1x + 76
Now plug in the value of x we just found

A = 2 + 76 = 78

8 0

3 years ago

3x7x7x7x7 in index notation 2x9x9x9x2x9 plz help thx

Leto [7]

$3 \times {7}^{4}$
Because there's one 3 & four 7's .

${2}^{2} \times {9}^{4}$
Because there are two 2's & four 9's

4 0

3 years ago

Read 2 more answers

Barb is building a shed 30 feet away from her house. On a map of her backyard, she drew the shed 14 inches away. Complete the sc

saul85 [17]

I think B 2.9 feet hope that helped

4 0

3 years ago

Read 2 more answers

A survey was conducted that asked 1003 people how many books they had read in the past year. Results indicated that x overbar eq

Sergio [31]

Answer:

The 99% confidence interval would be given (11.448;14.152).

Step-by-step explanation:

1) Important concepts and notation

A confidence interval for a mean "gives us a range of plausible values for the population mean. If a confidence interval does not include a particular value, we can say that it is not likely that the particular value is the true population mean"

$s=16.6$ represent the sample deviation

$\bar X=12.8$ represent the sample mean

n =1003 is the sample size selected

Confidence =99% or 0.99

$\alpha=1-0.99=0.01$ represent the significance level.

2) Solution to the problem

The confidence interval for the mean would be given by this formula

$\bar X \pm z_{\alpha/2} \frac{s}{\sqrt{n}}$

We can use a z quantile instead of t since the sample size is large enough.

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$12.8 - 2.58 \frac{16.6}{\sqrt{1003}}=11.448$

$12.8 + 2.58 \frac{16.6}{\sqrt{1003}} =14.152$

And the 99% confidence interval would be given (11.448;14.152).

We are confident that about 11 to 14 are the number of books that the people had read on the last year on average, at 1% of significance.

3 0

3 years ago