What is the unit price if 4 pounds of fruit cost $6.48

What is six ninths minus one third

san4es73 [151]

Answer: 1/3

Step-by-step explanation:

4 0

3 years ago

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Nara multiplied two whole numbers together to equal a product of 90.

BabaBlast [244]

Answer:

30×3=90 10×9=90 45×2=90 . .

4 0

3 years ago

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(CO6) From a random sample of 68 businesses, it is found that the mean time that employees spend on personal issues each week is

satela [25.4K]

Answer:

(1) (4.82, 4.98)

(2) Large sample size

(3) Yes, the temperature is within the confidence interval of (37.40, 37.60)

(4) (15.083, 15.117)

Step-by-step explanation:

Confidence Interval (CI) = mean + or - (t×sd)/√n

(1) mean = 4.9, sd = 0.35, n = 68, degree of freedom = n-1 = 68 - 1 = 67

t-value corresponding to 67 degrees of freedom and 95% confidence level is 1.9958

CI = 4.9 + (1.9958×0.35)/√68 = 4.98

CI = 4.9 - (1.9958×0.35)/√68 = 4.82

CI is (4.82, 4.98)

(2) Error margin = (t-value × standard deviation)/√sample size

From the formula above, error margin varies inversely as the square root of the sample size. Since the relationship between the error margin and sample size is inverse, increase in one (sample size) will conversely lead to a decrease in the other (error margin)

(3) mean = 37.5, sd = 0.6, n= 100, degree of freedom = n-1 = 100-1 = 99

t-value corresponding to 99 degrees of freedom and 90% confidence level is 1.6602

CI = 37.5 + (1.6602×0.6)/√100 = 37.5 + 0.10 = 37.60

CI = 37.5 - (1.6602×0.6)/√100 = 37.5 - 0.10 = 37.40

37.53 is within the confidence interval (37.40, 37.60)

(4) mean = 15.10, sd =0.08, n = 104, degree of freedom = n-1 = 104-1 = 103

t-value corresponding to 103 degrees of freedom and 97% confidence interval is 2.2006

CI = 15.10 + (2.2006×0.08)/√104 = 15.10 + 0.017 = 15.117

CI = 15.10 - (2.2006×0.08)/√104 = 15.10 - 0.017 = 15.083

CI is (15.083, 15.117)

4 0

3 years ago

What is b1 and r of geometrical sequence if b3 −b1 = 16 and b5 −b3 = 144.

Naily [24]

Answer:

b1 = 2 ; r = 3

Step-by-step explanation:

Given that :

if b3 −b1 = 16 and b5 −b3 = 144.

For a geometric series :

Ist term = a

Second term = ar

3rd term = ar^2

4th term = ar^3

5th term = ar^4 ;...

If b3 - b1 = 16;

ar^2 - a = 16

a(r^2 - 1) = 16 - - - (1)

b5 - b3 = 144

ar^4 - ar^2 = 144

ar^2(r^2 - 1) = 144 - - - - (2)

Divide (1) by (2)

a(r^2 - 1) / ar^2(r^2 - 1) = 16 /144

a / ar^2 = 1 / 9

ar^2 = 9a

Substitute for a in ar^2 - a = 16

9a - a = 16

8a = 16

a = 2

From ar^2 - a = 16

2r^2 - 2 = 16

2r^2 = 16 + 2

2r^2 = 18

r^2 = 18 / 2

r^2 = 9

r = √9

r = 3

Hence ;

a = b1 = 2 ; r = 3

7 0

3 years ago

A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x overbar, is found

joja [24]

Answer:

(a) 80% confidence interval for the population mean is [109.24 , 116.76].

(b) 80% confidence interval for the population mean is [109.86 , 116.14].

(c) 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed.

Step-by-step explanation:

We are given that a simple random sample of size n is drawn from a population that is normally distributed.

The sample mean is found to be 113 and the sample standard deviation is found to be 10.

(a) The sample size given is n = 13.

Firstly, the pivotal quantity for 80% 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 mean = 113

s = sample standard deviation = 10

n = sample size = 13

$\mu$ = population mean

Here for constructing 80% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 80% confidence interval for the population mean, $\mu$ is ;

P(-1.356 < $t_1_2$ < 1.356) = 0.80 {As the critical value of t at 12 degree

of freedom are -1.356 & 1.356 with P = 10%}

P(-1.356 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.356) = 0.80

P( $-1.356 \times }{\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

80% confidence interval for $\mu$ = [ $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.356 \times }{\frac{10}{\sqrt{13} } }$ , $113+1.356 \times }{\frac{10}{\sqrt{13} } }$ ]

= [109.24 , 116.76]

Therefore, 80% confidence interval for the population mean is [109.24 , 116.76].

(b) Now, the sample size has been changed to 18, i.e; n = 18.

So, the critical values of t at 17 degree of freedom would now be -1.333 & 1.333 with P = 10%.

80% confidence interval for $\mu$ = [ $\bar X-1.333 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.333 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.333 \times }{\frac{10}{\sqrt{18} } }$ , $113+1.333 \times }{\frac{10}{\sqrt{18} } }$ ]

= [109.86 , 116.14]

Therefore, 80% confidence interval for the population mean is [109.86 , 116.14].

(c) Now, we have to construct 98% confidence interval with sample size, n = 13.

So, the critical values of t at 12 degree of freedom would now be -2.681 & 2.681 with P = 1%.

98% confidence interval for $\mu$ = [ $\bar X-2.681 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+2.681 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-2.681 \times }{\frac{10}{\sqrt{13} } }$ , $113+2.681 \times }{\frac{10}{\sqrt{13} } }$ ]

= [105.56 , 120.44]

Therefore, 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed because t test statistics is used only when the data follows normal distribution.

6 0

3 years ago