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podryga [215]
3 years ago
6

If x=2, y=3 and z=1 then find the value of 1/2 x + 3/4 y

Mathematics
2 answers:
Nuetrik [128]3 years ago
7 0

Answer:

1 1/4

Step-by-step explanation:

x=2, y=3, and z=1

1/2(2)+3/4(3)=1+3/12=1 1/4

Anna71 [15]3 years ago
5 0

Answer:

decimal = 3.25

fraction = \frac{13}{4}

Step-by-step explanation:

since x and y are the only 2 variables being expressed in the question then it makes "z" irrelevant.

Therefore, you just need to plug in x and y into your question

(\frac{1}{2} × (2)) +( \frac{3}{4} × (3)) = 1 + 2.25 = 3.25 or \frac{13}{4}

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⦁ In a simple random sample of 1219 US adults, 354 said that their favorite sport to watch is football. Construct a 95% confiden
fomenos

Answer:

95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is [0.265 , 0.316].

Step-by-step explanation:

We are given that in a simple random sample of 1219 US adults, 354 said that their favorite sport to watch is football.

Firstly, the pivotal quantity for 95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is given by;

        P.Q. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } ~ N(0,1)

where, \hat p = proportion of adults in the United States whose favorite sport to watch is football in a sample of 1219 adults = \frac{354}{1219}

           n = sample of US adults  = 1291

           p = population proportion of adults

<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5%

                                                    significance level are -1.96 & 1.96}

P(-1.96 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } ) = 0.95

<u>95% confidence interval for p</u> = [ \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } , \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } ]

                         = [ \frac{354}{1219}-1.96 \times {\sqrt{\frac{\frac{354}{1219}(1-\frac{354}{1219})}{1219} } , \frac{354}{1219}+1.96 \times {\sqrt{\frac{\frac{354}{1219}(1-\frac{354}{1219})}{1219} } ]

                         = [0.265 , 0.316]

Therefore, 95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is [0.265 , 0.316].

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