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svlad2 [7]
3 years ago
12

Marietta is selling cheeses for the holiday fundraiser. On Monday, she sold 7/9 of the boxes of cheeses. On Tuesday she restocke

d her supply and sold 0.85 of the boxes of cheeses. On which day did Marietta sell more boxes? Explain. (HINT: Convert 7/9 into a decimal. To do this, take 7 and divide it by 9 then compare that decimal number with 0.85 and tell me which one is larger.)
Mathematics
2 answers:
leva [86]3 years ago
5 0
0.85 is larger than 7/9
Basile [38]3 years ago
5 0
0.85 is larger than 7/9

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May someone help me out with this please
yawa3891 [41]
60km  = 37.2823 

I don't know how many decimal places you are required to round but that is the precise answer.
3 0
3 years ago
The weights of 83 randomly selected windshields were found to have a variance of 1.88. Construct the 95% confidence interval for
Morgarella [4.7K]

Answer:

95% confidence interval for the population variance = (1.42 , 2.62).

Step-by-step explanation:

We are given that the weights of 83 randomly selected windshields were found to have a variance of 1.88.

<em>So, firstly the pivotal quantity for 95% confidence interval for the population variance is given by;</em>

        P.Q. = \frac{(n-1)s^{2} }{\sigma^{2} } ~ \chi^{2} __n_-_1

where, s^{2} = sample variance = 1.88

           \sigma^{2} = population variance

            n = sample of windshields = 83

So, 95% confidence interval for population variance, \sigma^{2} is;

P(58.85 < \chi^{2} __8_2 < 108.9) = 0.95 {As the table of \chi^{2} at 82 degree of freedom

                                              gives critical values of 58.85 & 108.9}

P(58.85 < \frac{(n-1)s^{2} }{\sigma^{2} } < 108.9) = 0.95

P( \frac{ 58.85}{(n-1)s^{2} } < \frac{1 }{\sigma^{2} } < \frac{108.9}{(n-1)s^{2} } ) = 0.95

P( \frac{ (n-1)s^{2}}{108.9 } < \sigma^{2} < \frac{ (n-1)s^{2}}{58.85 } ) = 0.95

<em><u>95% confidence interval for</u></em> \sigma^{2} = ( \frac{ (n-1)s^{2}}{108.9 } , \frac{ (n-1)s^{2}}{58.85 } )

                                                  = ( \frac{ (83-1)\times 1.88}{108.9 } , \frac{ (83-1)\times 1.88}{58.85 } )

                                                  = (1.42 , 2.62)

Therefore, 95% confidence interval for the population variance of the weights of all windshields in this factory is (1.42 , 2.62).

8 0
3 years ago
Find the value of x. Round to the nearest degree
ValentinkaMS [17]

Answer:

b. 44°

Step-by-step explanation:

Reference angle = x

Hypotenuse = 21

Adjacent = 15

Apply trigonometric function CAH:

Cos(x) = \frac{Adj}{Hyp}

Cos(x) = \frac{15}{21}

x = cos^{-1}(\frac{15}{21})

x = 44.4153086° ≈ 44° (nearest degree)

3 0
3 years ago
Given an infinite population that does not meet the conditions of a normal distribution, samples taken at random will also not b
AVprozaik [17]

Answer:

100% false

Reading the answer at this very moment



6 0
3 years ago
Read 2 more answers
Find the measure of angle COD.
8090 [49]

Answer:

45 degrees

Step-by-step explanation:

The value of COD is the angle between C and D

The value of C is 30

The value of D is 115

The difference is 115 -30 = 85

6 0
3 years ago
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