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2 years ago

12

In quadrilateral QRST, m is 68°, m is (3x + 40)°, and m is (5x − 52)°. What are the measures of , , and ? Write the numerical va

lues in that order with the measures separated by commas.

1 answer:

GrogVix [38]2 years ago

6 0

Answer:112,112,68

Step-by-step explanation:

You have the answer in the picture. You only need S and is 180-Q =180-68= 112, so the answer is 112,112,68

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Which monomial is a perfect cube?<br><br> 16x6<br> 27x8<br> 32x12<br> 64x6

Svetlanka [38]

Which monomial is a perfect cube?<span>16x6</span><span>27x8</span><span>32x12</span><span>64x<span>6

Its D </span></span>

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2 years ago

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The price of gasoline is $2.27 per gallon and decreases at the rate of $0.03 every two months. Oil costs $0.45 per gallon and in

kompoz [17]

You subtract 3 cents and add up how many months it is tiil it reaches the same ammout as adding 5 cents and adding up six months every time

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2 years ago

A certain type of thread is manufactured with a mean tensile strength of 78.3 kilograms and a standard deviation of 5.6 kilogram

azamat

Answer:

(a) The variance decreases.

(b) The variance increases.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Then, the mean of the sample mean is given by,

$\mu_{\bar x}=\mu$

And the standard deviation of the sample mean is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

The standard deviation of sample mean is inversely proportional to the sample size, <em>n</em>.

So, if <em>n</em> increases then the standard deviation will decrease and vice-versa.

(a)

The sample size is increased from 64 to 196.

As mentioned above, if the sample size is increased then the standard deviation will decrease.

So, on increasing the value of <em>n</em> from 64 to 196, the standard deviation of the sample mean will decrease.

The standard deviation of the sample mean for <em>n</em> = 64 is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{64}}=0.7$

The standard deviation of the sample mean for <em>n</em> = 196 is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{196}}=0.4$

The standard deviation of the sample mean decreased from 0.7 to 0.4 when <em>n</em> is increased from 64 to 196.

Hence, the variance also decreases.

(b)

If the sample size is decreased then the standard deviation will increase.

So, on decreasing the value of <em>n</em> from 784 to 49, the standard deviation of the sample mean will increase.

The standard deviation of the sample mean for <em>n</em> = 784 is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{784}}=0.2$

The standard deviation of the sample mean for <em>n</em> = 49 is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{49}}=0.8$

The standard deviation of the sample mean increased from 0.2 to 0.8 when <em>n</em> is decreased from 784 to 49.

Hence, the variance also increases.

6 0

2 years ago

Your company has just taken out a 1-year installment loan for $72,500 at a nominal rate of 20.0% but with equal end-of-month pay

Mariulka [41]

Answer:

100% of the 2nd monthly payment go toward the repayment of principal.

Step-by-step explanation:

The loan taken is the Principal which is mentioned as $72,500 with interest at a nominal rate of 20%. Firstly, it is important to understand that nominal rate means <em>non-compounding </em>rate. Simply put will be a "<em>one-time charged" </em>rate on the loan. Since this is given as 20% of the Principal. It is calculated thus: $\frac{20}{100}$ × $\frac{72,500}{1}$ = $14,500. So the interest on the loan is $14,500. Added to the Principal the total amount to be paid back by the company becomes: $72,500 + $14,500 = $87,000. To pay back this amount at equal end-of-month installments in 1 year (12 months), we divide the total amount by 12. i.e $\frac{87000}{12}$ = $7250. This means, the monthly payment will be $7,250. Since the monthly payment pays only 10% of the initial principal $72,500. By the second month only 20% of the Principal would have been paid. So all of the monthly payment will go towards repaying the principal

3 0

2 years ago

Find each measure for the given set of data: 11, 13, 17, 20, 22, 25, 27, 31, 31, 33, Mean= Median= Range= interquartile range= I

Tresset [83]

Answer:

Mean=23

Median =23.5

Range=22

Interquartile range = 14

Step-by-step explanation:

Given set of data

11,13,17,20,22,25,27,31,31,33

Mean $=\frac{11+13+17+20+22+25+27+31+31+33}{10}$

$=\frac{230}{10}$

=23

The $(\frac{n}{2})$ th term= the 5th term

=22

The $(\frac{n}{2}+1)^{th}$ term = The $6^{th$ term

= 25

The median = $\frac{22+25}{2}$

=23.5

Range = Highest term - lowest term

=33- 11

=22

Here lower half {11,13,17,20,22}

The middle number of lower half is first quartile.

Q₁ = 17

And lower half is{25,27,31,31,33}

The middle number of upper half is third quartile.

Q₃=31

Interquartile Range =Q₃-Q₁

=31-17

=14

8 0

2 years ago