Joe read 140 of his 300 page book. What percent of the book has he read

Robert read 20 pages out of a 300 page book. What percent has he read ? What percent does he still have to read ?

Setler [38]

Answer: he has read 15% and still needs to read 85%

Step-by-step explanation: 300/20 is 15 so that’s the percent he has read. 100-15=85 so that’s the percent he still has left to read

6 0

2 years ago

A survey was conducted to determine the average age at which college seniors hope to retire in a simple random sample of 101 sen

tatyana61 [14]

Answer:

96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].

Step-by-step explanation:

We are given that a survey was conducted to determine the average age at which college seniors hope to retire in a simple random sample of 101 seniors, 55 was the average desired retirement age, with a standard deviation of 3.4 years.

Firstly, the Pivotal quantity for 96% 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 average desired retirement age = 55 years

$\sigma$ = sample standard deviation = 3.4 years

n = sample of seniors = 101

$\mu$ = true mean retirement age of all college students

Here for constructing 96% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 96% confidence interval for the population mean, $\mu$ is ;

P(-2.114 < $t_1_0_0$ < 2.114) = 0.96 {As the critical value of t at 100 degree

of freedom are -2.114 & 2.114 with P = 2%}

P(-2.114 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.114) = 0.96

P( $-2.114 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.114 \times {\frac{s}{\sqrt{n} } }$ ) = 0.96

P( $\bar X-2.114 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.114 \times {\frac{s}{\sqrt{n} } }$ ) = 0.96

96% confidence interval for $\mu$ = [ $\bar X-2.114 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.114 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $55-2.114 \times {\frac{3.4}{\sqrt{101} } }$ , $55+2.114 \times {\frac{3.4}{\sqrt{101} } }$ ]

= [54.30 , 55.70]

Therefore, 96% confidence interval for desired retirement age of all college students is [54.30 , 55.70].

7 0

2 years ago

What is the GCF 16 24 40

boyakko [2]

The factors of 16 are 1, 2, 4, 8, and 16.
The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24. 
The factors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40. 
So the greatest common factor of 16, 24, and 40 is 8.

Hope this helps:)

4 0

2 years ago

Multiply. (2x2 – 3x + 1)(x2 - 4x – 3)

Arisa [49]

Answer:

2x^4−11x^3+7x^2+5x−3

Step-by-step explanation:

The ^ means exponent

3 0

2 years ago

I am greater than 15 I am less than 21 I have 8 ones

PSYCHO15rus [73]

Answer:

18

Step-by-step explanation:

bc 8 digits in the ones place, and less than 21 greater than 15

7 0

3 years ago

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