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Whitepunk [10]
2 years ago
13

How many days are there in 4 weeks,3 days?

Mathematics
2 answers:
Novay_Z [31]2 years ago
7 0
31 days because 4 times 7 = 28 + 3 =31
dybincka [34]2 years ago
6 0
The answer is 31 days because if you multiply 4 and 7 you get 28. Then you add 3 to 28, and you get 31 days.
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The graph shown below is a scatter plot:
Verdich [7]
I don’t see a point so how am I supposed to answer
8 0
2 years ago
Look at the figure what is the value of y √2y+3=11
Natasha_Volkova [10]

Answer:

y=2

Step-by-step explanation:

first solve those in the brackets then find the square root

2y+3=11

put like terms together

3 goes to the right side

2y=11-3

2=8

y=4

√4=2

y=2

3 0
3 years ago
A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past
Vadim26 [7]

Answer:

We conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.

Step-by-step explanation:

We are given that a particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8000.

From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9800 miles.

<u><em>Let </em></u>\mu<u><em> = average miles for deluxe tires</em></u>

So, Null Hypothesis, H_0 : \mu \geq 50,000 miles   {means that deluxe tire averages at least 50,000 miles before it needs to be replaced}

Alternate Hypothesis, H_A : \mu < 50,000 miles    {means that deluxe tire averages less than 50,000 miles before it needs to be replaced}

The test statistics that will be used here is <u>One-sample z test statistics</u> as we know about population standard deviation;

                                  T.S.  = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean lifespan = 46,500 miles

            \sigma = population standard deviation = 8000 miles

            n = sample of tires = 28

So, <u><em>test statistics</em></u>  =  \frac{46,500-50,000}{\frac{8000}{\sqrt{28} } }

                               =  -2.315

The value of the test statistics is -2.315.

Now at 5% significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is less than the critical value of z as -2.315 < -1.6449, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.

4 0
3 years ago
Suppose 52% of the population has a college degree. If a random sample of size 563563 is selected, what is the probability that
amm1812

Answer:

The value is  P(| \^ p -  p| < 0.05 ) = 0.9822

Step-by-step explanation:

From the question we are told that

    The population proportion is  p =  0.52

     The sample size is  n  =  563      

Generally the population mean of the sampling distribution is mathematically  represented as

           \mu_{x} =  p =  0.52

Generally the standard deviation of the sampling distribution is mathematically  evaluated as

       \sigma  =  \sqrt{\frac{ p(1- p)}{n} }

=>      \sigma  =  \sqrt{\frac{ 0.52 (1- 0.52 )}{563} }

=>      \sigma  =   0.02106

Generally the  probability that the proportion of persons with a college degree will differ from the population proportion by less than 5% is mathematically represented as

            P(| \^ p -  p| < 0.05 ) =  P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 ))

  Here  \^ p is the sample proportion  of persons with a college degree.

So

 P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P(\frac{[[0.05 -0.52]]- 0.52}{0.02106} < \frac{[\^p - p] - p}{\sigma }  < \frac{[[0.05 -0.52]] + 0.52}{0.02106} )

Here  

    \frac{[\^p - p] - p}{\sigma }  = Z (The\ standardized \  value \  of\  (\^ p - p))

=> P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P[\frac{-0.47 - 0.52}{0.02106 }  <  Z  < \frac{-0.47 + 0.52}{0.02106 }]

=> P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P[ -2.37 <  Z  < 2.37 ]

=>  P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = P(Z <  2.37 ) - P(Z < -2.37 )

From the z-table  the probability of  (Z <  2.37 ) and  (Z < -2.37 ) is

  P(Z <  2.37 ) = 0.9911

and

  P(Z <  - 2.37 ) = 0.0089

So

=>P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) =0.9911-0.0089

=>P( - (0.05 - 0.52 ) <  \^ p <  (0.05 + 0.52 )) = 0.9822

=> P(| \^ p -  p| < 0.05 ) = 0.9822

3 0
3 years ago
Select the best answer for the question.
fgiga [73]
The answer is D - 4/5
7 0
2 years ago
Read 2 more answers
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