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2 years ago

Find the surface area of the solid figure. The surface area of the figure is (use whole numbers only):

Mathematics

1 answer:

fgiga [73]2 years ago

6 0

Answer:

Hello! answer: the area is 528

Step-by-step explanation:

When finding area you should split the shapes up so forst im going to do the two sides that are the same those are just simply 10 x 12 = 120 10 x 12 = 120 So i add those up and get 240 I'll add this with the other sides later now Im going to do the front face thats a square that is 16 x 12 = 192 Now I'm just going to do the triangles that is 16 x 6 = 96 then I divide by 2 cause thats the formula for finding triangle area now I have 48 but since there are two of those triangles ill just do 48 x 2 = 96 now that I have All the faces im going to add them all up so... 240+ 192 + 96 = 528 Hope this explains well!

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Solve for x <br>81^5 = 3x<br>x=?<br>

pentagon [3]

81^5 = 3,486,784,401
3,486,784,401 / = X
X = 1,162,261,467
(I think anyways)

8 0

2 years ago

janelle wishes to finance a car for $33,000. the bank's annual interest rate is 3.5%, and she can choose between durations of fi

Harman [31]

Answer:

For a duration of 5 years, Monthly Payment =$600.42

For a duration of 6 years, Monthly Payment =$508.83

Step-by-step explanation:

$P=\dfrac{ar (1+r)^n}{(1+r)^n-1} \\$

where a= Amount to Finance=$33,000

Annual interest rate = 3.5%=0.035

r=Monthly Interest Rate= 0.035 ÷ 12 = $\frac{7}{2400}$

n=number of months to pay

For a duration of 5 years

n=5X12=60 months

$P=\dfrac{ar (1+r)^n}{(1+r)^n-1} \\\\P=\dfrac{33000 X\frac{7}{2400} (1+\frac{7}{2400} )^{60}}{(1+\frac{7}{2400})^{60}-1} \\=\dfrac{96.25 (1.1909)}{1.1909-1}\\=\dfrac{96.25 (1.1909)}{0.1909}\\=\dfrac{114.62}{0.1909}=\$600.42$

For a duration of 6 years

n=6X12=72 months

$P=\dfrac{ar (1+r)^n}{(1+r)^n-1} \\\\P=\dfrac{33000 X\frac{7}{2400} (1+\frac{7}{2400} )^{72}}{(1+\frac{7}{2400})^{72}-1} \\=\dfrac{96.25 (1.2333)}{1.2333-1}\\=\dfrac{96.25 (1.2333)}{0.2333}\\=\dfrac{118.71}{0.2333}=\$508.83$

4 0

2 years ago

There are two games involving flipping a coin. In the first game you win a prize if you can throw between 45% and 55% heads. In

Nina [5.8K]

Answer:

d) 300 times for the first game and 30 times for the second

Step-by-step explanation:

We start by noting that the coin is fair and the flip of a coin has a probability of 0.5 of getting heads.

As the coin is flipped more than one time and calculated the proportion, we have to use the <em>sampling distribution of the sampling proportions</em>.

The mean and standard deviation of this sampling distribution is:

$\mu_p=p\\\\ \sigma_p=\sqrt{\dfrac{p(1-p)}{N}}$

We will perform an analyisis for the first game, where we win the game if the proportion is between 45% and 55%.

The probability of getting a proportion within this interval can be calculated as:

$P(0.45$

referring the z values to the z-score of the standard normal distirbution.

We can calculate this values of z as:

$z_H=\dfrac{p_H-\mu_p}{\sigma_p}=\dfrac{(p_H-p)}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p_H-p)>0\\\\\\z_L=\dfrac{p_L-\mu_p}{\sigma_p}=\dfrac{p_L-p}{\sqrt{\dfrac{p(1-p)}{N}}}=\sqrt{\dfrac{N}{p(1-p)}}*(p_L-p)$

If we take into account the z values, we notice that the interval increases with the number of trials, and so does the probability of getting a value within this interval.

With this information, our chances of winning increase with the number of trials. We prefer for this game the option of 300 games.

For the second game, we win if we get a proportion over 80%.

The probability of winning is:

$P(p>0.8)=P(z>z^*)$