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SIZIF [17.4K]
3 years ago
8

The height of a triangle is 8 m less than the base. The area is 10 m2. Find the height and base.​

Mathematics
1 answer:
Lana71 [14]3 years ago
8 0

Answer:

he height of a triangle is 8m less than the base. The area is 10cm^2. Find the height and the base.

I have tried the equation 10=b(b-8)(1/2).

b(b-8) = 20

b^2 - 8b - 20 = 0

Factor:

b^2-10b + 2b-20 = 0

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A toy company makes rectangular sandboxes that measure 6 feet by 5 feet by 1.2 feet. A customer buys a sandbox and 40 cubic feet
Iteru [2.4K]

Answer:

Customer has bought too much sand

Step-by-step explanation:

Given that the dimensions (Length, Width and Height) of rectangular box are = 6,5, 1.2.

Volume of a rectangular box = Length * Width * Height

=> 6*5*1.2

=> 36 cubic feet

As mentioned, customer bought a sand box and <u>40 cubic feet of sand.</u>

So,

Volume of rectangular box (R) = 36 cubic feet

Volume of Sand (S) = 40 cubic feet

From analysis

S > R

Which shows that the capacity of sandbox is 36 cubic feet and volume is 40 feet that is a little greater than desired capacity.

Therefore, customer has bought too much sand.

4 0
3 years ago
In 16% of all homes with a stay-at-home parent, the father is the stay-at-home parent (Pew Research, June 5, 2014). An independe
enot [183]

Answer:

a)

n = (\frac{z\sqrt{0.16*0.84}}{0.03})^2, in which z is related to the confidence level.

b) A sample size of 991 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

In 16% of all homes with a stay-at-home parent, the father is the stay-at-home parent

This means that \pi = 0.16

a. What sample size is needed if the research firm's goal is to estimate the current proportion of homes with a stay-at-home parent in which the father is the stay-at-home parent with a margin of error of 0.03 (round up to the next whole number).

This is n for which M = 0.03. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.03 = z\sqrt{\frac{0.16*0.84}{n}}

0.03\sqrt{n} = z\sqrt{0.16*0.84}

\sqrt{n} = \frac{z\sqrt{0.16*0.84}}{0.03}

(\sqrt{n})^2 = (\frac{z\sqrt{0.16*0.84}}{0.03})^2

n = (\frac{z\sqrt{0.16*0.84}}{0.03})^2, in which z is related to the confidence level.

Question b:

99% confidence level,

So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

n = (\frac{z\sqrt{0.16*0.84}}{0.03})^2

n = (\frac{2.575\sqrt{0.16*0.84}}{0.03})^2

n = 990.2

Rounding up

A sample size of 991 is needed.

8 0
3 years ago
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