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creativ13 [48]
3 years ago
8

What is the gcf of 68,76 and 99?

Mathematics
1 answer:
BabaBlast [244]3 years ago
5 0
68 = 2 x 2 x 17
76 = 2 x 2 x 19
99 = 3 x 3 x 11

gcf = 1
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Hi can you please help me​
astra-53 [7]

Answer:

<h2>Area = 90 cm²</h2><h2>Perimeter = 53 cm</h2>

Step-by-step explanation:

area and perimeter of the given diagram

<u>area = 1/2 * base * height</u>

where base = 15 cm

        height = 12 cm

plugin values into the formula

area = 1/2 * 15 * 12

area = 90 cm²

<u>Perimeter = add up all sides (AB + AC + BC)</u>

where side BC = √(5² + 12²)

                   BC = 13 cm

Perimeter = 25cm + 15 cm + 13 cm

Perimeter = 53 cm

6 0
3 years ago
Electricity usage data consists of 45 months has a mean number of units consumed is 390.47 per month with a standard deviation o
neonofarm [45]

Answer:

The 95% confidence interval for the average monthly electricity consumed units is between 47.07 and 733.87

Step-by-step explanation:

We have the standard deviation for the sample. So we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 45 - 1 = 44

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 44 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.95}{2} = 0.975. So we have T = 2.0141

The margin of error is:

M = T*s = 2.0141*170.5 = 343.4

In which s is the standard deviation of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 390.47 - 343.40 = 47.07 units per month

The upper end of the interval is the sample mean added to M. So it is 390.47 + 343.40 = 733.87 units per month

The 95% confidence interval for the average monthly electricity consumed units is between 47.07 and 733.87

6 0
3 years ago
F (x) = 3x + 7 <br> F (-4) = <br> F (2) =
V125BC [204]

Answer:

f(-4)=3(-4)+7=0

f(2)=3(2)+7=13

3 0
3 years ago
Read 2 more answers
If we are given x such that 25^x-9^y=18 and 5^x-3^y=3, compute 5^x+3^y.
leonid [27]

25 = 5*5 and 9 = 3*3, which we can exploit to write

25^x-9^y=(5^2)^x-(3^2)^y=5^{2x}-3^{2y}=(5^x)^2-(3^y)^2

so that this expression is actually a difference of squares. We can factorize this to get

(5^x-3^y)(5^x+3^y)=18

and given that 5^x-3^y=3, we divide both sides by this to get

5^x+3^y=\dfrac{18}3=\boxed{6}

7 0
3 years ago
It is almost the end of the grading period. I have 730 out of 880 points so far in my math class which means I currently have a
maksim [4K]
What is being asked here? Are they asking for the passing grade or the total with all the other 90 points added? I think it is A, 90%
7 0
3 years ago
Read 2 more answers
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