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2 years ago

7. If SK = 13x - 5, KY= 2x + 9, and SY = 36-x, find each value.

Mathematics

1 answer:

sweet [91]2 years ago

5 0

Answer:

SY=34, SK=21 and KY=13

Step-by-step explanation:

we have that

SY=SK+KY

substitute the given values

(36-x)=(13x-5)+(2x+9)

solve for x

36-x=15x+4

15x+x=36-4

16x=32

x=2

Find the value of SY

SY=(36-x)=36-2=34

Find the value of SK

SK=(13x-5)=13(2)-5=21

Find the value of KY

KY=(2x+9)=2(2)+9=13

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A random sample of 250 students at a university finds that these students take a mean of 15.2 credit hours per quarter with a st

Orlov [11]

Answer:

95% confidence interval for the mean credit hours taken by a student each quarter is [14.915 hours , 15.485 hours].

Step-by-step explanation:

We are given that a random sample of 250 students at a university finds that these students take a mean of 15.2 credit hours per quarter with a standard deviation of 2.3 credit hours.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample credit hours per quarter = 15.2 credit hours

s = sample standard deviation = 2.3 credit hours

n = sample of students = 250

$\mu$ = population mean credit hours per quarter

Here for constructing 95% confidence interval we have used One-sample t test statistics as we know don't about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < $t_2_4_9$ < 1.96) = 0.95 {As the critical value of t at 249 degree of

freedom are -1.96 & 1.96 with P = 2.5%}

P(-1.96 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.96 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $15.2-1.96 \times {\frac{2.3}{\sqrt{250} } }$ , $15.2+1.96 \times {\frac{2.3}{\sqrt{250} } }$ ]

= [14.915 hours , 15.485 hours]

Therefore, 95% confidence interval for the mean credit hours taken by a student each quarter is [14.915 hours , 15.485 hours].

The interpretation of the above confidence interval is that we are 95% confident that the true mean credit hours taken by a student each quarter will be between 14.915 credit hours and 15.485 credit hours.

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3.52 A coin is tossed twice. Let Z denote the number of heads on the first toss and W the total number of heads on the 2 tosses.

Rufina [12.5K]

Answer:

a) The joint probability distribution

P(0,0) = 0.36, P(1,0) = 0.24, P(2,0) = 0, P(0,1) = 0, P(1,1) = 0.24, P(2,1)= 0.16

b) P( W = 0 ) = 0.36, P(W = 1 ) = 0.48, P(W = 2 ) = 0.16

c) P ( z = 0 ) = 0.6

P ( z = 1 ) = 0.4

Step-by-step explanation:

Number of head on first toss = Z

Total Number of heads on 2 tosses = W

% of head occurring = 40%

% of tail occurring = 60%

P ( head ) = 2/5 , P( tail ) = 3/5

a) Determine the joint probability distribution of W and Z

P( W =0 |Z = 0 ) = 0.6 P( W = 0 | Z = 1 ) = 0

P( W = 1 | Z = 0 ) = 0.4 P( W = 1 | Z = 1 ) = 0.6

P( W = 1 | Z = 0 ) = 0 P( W = 2 | Z = 1 ) = 0.4

The joint probability distribution

P(0,0) = 0.36, P(1,0) = 0.24, P(2,0) = 0, P(0,1) = 0, P(1,1) = 0.24, P(2,1)= 0.16

B) Marginal distribution of W

P( W = 0 ) = 0.36, P(W = 1 ) = 0.48, P(W = 2 ) = 0.16

C) Marginal distribution of Z ( pmf of Z )

P ( z = 0 ) = 0.6

P ( z = 1 ) = 0.4

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2 years ago

Please consider the following data set. The initial row provides labels for two variables (i.e., X1 and X2). Each subsequent row

LekaFEV [45]

Answer:

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Step-by-step explanation:

Given is the data set for X1 and x2 are we have to do a paired t test.

$H_0: \bar X_1 = \bar X_2\\H_a: \bar X_1 \neq \bar X_2$

(Two tailed test at 5% significance level)

Mean difference = 1.00

Std error for difference = 1.225

df=4

Test statistic = mean diff/se = 1.826

p value =0.142

Since p >0.05 our alpha we accept H0

There is no significant difference before and after averages at 5% sign. level.

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2 years ago