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cestrela7 [59]
2 years ago
7

Help :((( please will mark brainless​

Mathematics
1 answer:
nikklg [1K]2 years ago
5 0

Answer:

A_{total} = 25\ ft^2

Step-by-step explanation:

<u>Step 1:  Determine the area of the square</u>

A = s^2

A = (3\ ft)^2

A = 9\ ft^2

<u>Step 2:  Split the top figure into a square and triangle</u>

Determine the area of the rectangle

A = l * w

A = (7\ ft - 3\ ft) * (5\ ft - 2\ ft)

A = (4\ ft) * (3\ ft)

A = 12\ ft^2

Determine the area of the triangle

A = \frac{b * h}{2}

A = \frac{\frac{7\ ft - 3\ ft}{2} * (5\ ft - 3\ ft)}{2}

A=\frac{2\ ft * 2\ ft}{2}

A = \frac{4\ ft^2}{2}

A = 2\ ft^2

Since there are two right triangles in the triangle that we have we multiply the area that we got by 2 to get the total area.

A_{triangle} = 2\ ft^2*2

A_{triangle} = 4\ ft^2

<u>Step 3:  Determine the total area</u>

A_{total} = 9\ ft^2 + 12\ ft^2 + 4\ ft^2

A_{total} = 25\ ft^2

Answer:  A_{total} = 25\ ft^2

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s defined to be the dollar value of loans defaulted divided by the total dollar value of all loans made. Banking officials claim
Amiraneli [1.4K]

Answer:

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the mean bad debt ratio for Ohio banks is different than the Midwestern average (3.5%).

Test statistic t = 1.431

Critical values tc = ±2.447

P-value = 0.203

Step-by-step explanation:

This is a hypothesis t-test for the population mean.

The claim is that the mean bad debt ratio for Ohio banks is different than the Midwestern average (3.5%).

Then, the null and alternative hypothesis are:

H_0: \mu=3.5\\\\H_a:\mu\neq 3.5

The significance level is 0.05.

The sample has a size n=7.

We calculate  the sample mean and standard deviation as:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{7}(7+4+6+3+5+4+2)\\\\\\M=\dfrac{31}{7}\\\\\\M=4.43\\\\\\

s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{6}((7-4.43)^2+(4-4.43)^2+(6-4.43)^2+. . . +(2-4.43)^2)}\\\\\\s=\sqrt{\dfrac{17.71}{6}}\\\\\\s=\sqrt{2.95}=1.72\\\\\\

The sample mean is M=4.43.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=1.72.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{1.72}{\sqrt{7}}=0.65

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{4.43-3.5}{0.65}=\dfrac{0.93}{0.65}=1.431

The degrees of freedom for this sample size are:

df=n-1=7-1=6

This test is a two-tailed test, with 6 degrees of freedom and t=1.431, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=2\cdot P(t>1.431)=0.203

As the P-value (0.203) is bigger than the significance level (0.05), the effect is not significant.

If we use the critical value approach, for this level of confidence, the critical values are tc = ±2.447. The test statistic is within the bounds of the critical values and falls within the acceptance region.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the mean bad debt ratio for Ohio banks is different than the Midwestern average (3.5%).

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Solve cos x + rad 2 = -cos x for over the interval [0, 2pi ]
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cos x + \sqrt{2} = -cosx \\  \\ 2 cos x + \sqrt{2} = 0 \\  \\ cos x = -\frac{\sqrt{2}}{2}

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Refer to a unit circle to find that 
cos (\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2}  \\  \\ cos (\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2}

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