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inessss [21]
3 years ago
5

What is a area of a triangle with a heigt of 5 inches and a base of 10inches

Mathematics
1 answer:
nydimaria [60]3 years ago
7 0

Answer: 25 in²

Step-by-step explanation: To find the area of a triangle, start with the formula for the area of a triangle which is shown below.

Area = \frac{1}{2} bh

In this problem, we're given that the base is 10 inches

and the height is 5 inches.

Now, plugging into the formula, we have (\frac{1}{2})(10in.)(5 in.).

Now, it doesn't matter which order we multiply.

So we can begin by multiplying (1/2) (10 in.) to get 5 inches.

Now, (5 in.) (5 in.) is 25 in².

So the area of the triangle is 25 in².

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Use the empirical rule to solve the problem. The annual precipitation for one city is normally distributed with a mean of 288 in
nadezda [96]

Answer:

z=-1.99

z=1.99

And if we solve for a we got

a=288 -1.99*3.7=214.4

a=288 +1.99*3.7=295.4

And the limits for this case are: (214.4; 295.4)

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the annual precipitation of a population, and for this case we know the distribution for X is given by:

X \sim N(288,3.7)  

Where \mu=288 and \sigma=3.7

The confidence level is 95.44 and the signficance is 1-0.9544=0.0456 and the value of \alpha/2 =0.0228. And the critical value for this case is z = \pm 1.99

Using this condition we can find the limits

z=-1.99

z=1.99

And if we solve for a we got

a=288 -1.99*3.7=214.4

a=288 +1.99*3.7=295.4

And the limits for this case are: (214.4; 295.4)

8 0
3 years ago
Malik earned $2,344.55 fixing computers. He charged $35.50 per hour. Which expression does NOT use appropriate compatible number
Mademuasel [1]
I will say b 2,400 and some else
3 0
3 years ago
1. Fifty-eight thousandths
kap26 [50]

fifty-eight thousandths?

4 0
3 years ago
The length of each side of a regular pentagon is increased by 8 inches, so the perimeter is now 65 inches. What is the original
evablogger [386]

<em>Here</em> as the <em>Pentagon</em> is <em>regular</em> so it's <em>all sides</em> will be of <em>equal length</em> . And if we assume It's each side be<em> </em><em><u>s</u></em> , then it's perimeter is going to be <em>(s+s+s+s+s) = </em><em><u>5s</u></em>.And as here , each <em>side</em> is increased by <em>8 inches</em> and then it's perimeter is <em>65 inches</em> , so we got that it's side after increament is<em> (s+8) inches</em> and original length is <em>s inches </em>. And if it's each side is <em>(s+8) inches</em> , so it's perimeter will be <em>5(s+8)</em> and as it's equal to <em>65 inches</em> . So , <em><u>5(s+8) = 65</u></em>

{:\implies \quad \sf 5(s+8)=65}

{:\implies \quad \sf 5s+5\times 8=65}

{:\implies \quad \sf 5s+40=65}

{:\implies \quad \sf 5s=65-40}

{:\implies \quad \sf 5s=25}

{:\implies \quad \sf s=\dfrac{25}{5}=5}

{:\implies \quad \bf \therefore \quad \underline{\underline{s=5\:\: Inches}}}

As we assumed the original side to be <em><u>s</u></em> .

<em>Hence, the original side's length 5 inches </em>

8 0
2 years ago
A fair dice has six sides, numbered 1 to 6
Ivanshal [37]

Answer:

a. 1/6

Step-by-step explanation:

for a, there is only one side that has three

8 0
3 years ago
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