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Zepler [3.9K]
2 years ago
8

What is the area of the figure below​

Mathematics
1 answer:
laila [671]2 years ago
5 0

Answer:

Hello! answer: area = 21

Step-by-step explanation:

5 × 3 = 15 3 × 2 = 6 15 + 6 = 21 HOPE THAT HELPS!

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IM SOOOO STUCKKKK HELLPPPPPPPPPPP!!
KonstantinChe [14]

Answer:

18

Step-by-step explanation:

x² = 16² + 30² - 2(16)(30)cos 30°

x² = 256 + 900 - 960(0.866)

x² = 1156 - 831.36

x² = 324.64

x = 18.02

Approx. 18

Please mark my answer as brainliest

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3 years ago
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0.7 mm to centimeters
san4es73 [151]

Answer: 0.07 centimeters

Step-by-step explanation: 0.7millimeters equals 0.07 centimeters

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LAST QUESTION thank you so much to nintendo who has been helping me with all of my questions!
melomori [17]
Using the properties of arcs and inscribed angles, B is 110/2=55.
3 0
2 years ago
Alina says the area of the rectangular face is 10 in 2 ls
Mumz [18]

Answer:

It's B

Step-by-step explanation:

Because I randomly guessed and that was the answer.

5 0
2 years ago
Read 2 more answers
A sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.
TEA [102]

Answer:

(a) Null Hypothesis, H_0 : \mu = $1,150  

    Alternate Hypothesis, H_A : \mu \neq $1,150

(b) The test statistic is 3.571.

(c) We conclude that the mean of all account balances is significantly different from $1,150.

Step-by-step explanation:

We are given that a sample of 81 account balances of a credit company showed an average balance of $1,200 with a standard deviation of $126.

We have test the hypothesis to determine whether the mean of all account balances is significantly different from $1,150.

<u><em>Let </em></u>\mu<u><em> = mean of all account balances</em></u>

(a)So, Null Hypothesis, H_0 : \mu = $1,150     {means that the mean of all account balances is equal to $1,150}

Alternate Hypothesis, H_A : \mu \neq $1,150    {means that the mean of all account balances is significantly different from $1,150}

The test statistics that will be used here is <u>One-sample t test statistics</u> as we don't know about population standard deviation;

                               T.S.  = \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample average balance = $1,200

             s = sample standard deviation = $126

             n = sample of account balances = 81

(b) So, <u>test statistics</u>  =  \frac{1,200-1,150}{\frac{126}{\sqrt{81} } }  ~ t_8_0

                               =  3.571

The value of the sample test statistics is 3.571.

(c) <u>Now, P-value of the test statistics is given by the following formula;</u>

          P-value = P( t_8_0 > 3.571) = <u>Less than 0.05%</u>

Because in the t table the highest critical value for t at 80 degree of freedom is given between 3.460 and 3.373 at 0.05% level.

Now, since P-value is less than the level of significance as 5% > 0.05%, so we sufficient evidence to reject our null hypothesis.

Therefore, we conclude that the mean of all account balances is significantly different from $1,150.

8 0
3 years ago
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