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Murljashka [212]
3 years ago
7

What is the value of ?

Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
4 0
The value of the angle is 105 degrees.
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Can you create a triangle with sides 17ft , 8ft , and 25ft ?
Vlad1618 [11]

Answer:

Step-by-step explanation:

Yes

5 0
4 years ago
Read 2 more answers
Jamal has at most $100 to spend on computer paper and printer ink for his computer. He buys 1 box of paper for $24.99. If ink ca
lions [1.4K]

Answer:

1 box of paper with 3 ink cartridges

Step-by-step explanation:

$100 total

$24.99+19.99(3)= $84.96 (4 ink cartridges would be over the allotted amount)

7 0
3 years ago
Can you solve this for me please in a good way
zhannawk [14.2K]

Answer:

x = -2

Step-by-step explanation:

If you factor out 1/9 you get rid of the fractions and that makes the equation a whole lot easier.

(5x+10)^2 - x(25x+27) + 46 = 0

Next, eliminate the brackets:

25x^2 +100x + 100 - 25x^2 - 27x + 46 = 0

Combine the terms:

x = \frac{- 146}{73} = -2


3 0
3 years ago
Read 2 more answers
(b +11) (b - 11)<br><br> Please show work
Inessa05 [86]

Answer:

b² - 121

Step-by-step explanation:

(b +11)(b - 11)

b² - 11b + 11b - 121

b² - 121

5 0
3 years ago
The bad debt ratio for a financial institution is defined to the dollar value of loans defaulted divided by the total dollar val
vlabodo [156]

Answer:

Step-by-step explanation:

From the information given:

Consider X to be the random variable denoting the bad debt ratios for Ohio Bank.

Then, X \sim N ( \mu, \sigma ^2)

Thus the null hypothesis and the alternative can be computed as:

Null hypothesis:

H_o :  \mu \leq3.5\%

Alternative hypothesis

H_1 : \mu > 3.5\%

The type I and type II error is as follows:

Type I:

The mean bad debt ratio is > 3.5% when it is not

Type II:

The mean bad debt ratio is ≤ 3.5% when it is not.

The test statistics can be calculated by using the formula:

t = \dfrac{\overline x - \mu}{\dfrac{\sigma}{\sqrt{n}}}

where;

sample size n = 7

mean = 6+8+5+9+7+5+8 = 48

sample mean \overline x =\dfrac{48}{7}

\overline x = 6.86

sample standard deviation is :

s = \sqrt{\dfrac{\sum( x -\overline x)^2}{n-1}}

s = \sqrt{\dfrac{( 6 -6.86)^2+( 8-6.86)^2+  ( 5 -6.86)^2 + ...+( 7 -6.86)^2+ ( 5 -6.86)^2+( 8 -6.86)^2 }{7-1}}

s = 1.573

population  mean \mu = 3.5

Therefore, the test statistics is :

t = \dfrac{\overline x - \mu}{\dfrac{\sigma}{\sqrt{n}}}

t = \dfrac{6.86- 3.5}{\dfrac{1.573}{\sqrt{7}}}

t = \dfrac{3.36}{\dfrac{1.573}{2.65}}

t = \dfrac{3.36\times {2.65}}{{1.573}}

t = 5.660

At significance level of 0.01

t_{0.01} = 3.707  

P - value = P(T > 5.66)

P - value = 1 - (T < 5.66)

P - value = 1 - 0.9993

P-value = 0.0007

Therefore, since t_{0.01} < t , we reject the null hypothesis and conclude that the claim that the mean bad debt ratio for Ohio banks is higher than the mean for all financial institutions is true.

4 0
4 years ago
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