The difference of the squares of two consecutive odd integers is 24. Find the integers.

There are a total of 91 students in a drama club and yearbook club. The drama club has 19 more students than the yearbook club.

jasenka [17]

91+ 19
Answer: 110
This is what the answer is :)

7 0

3 years ago

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Tina is placing 30 roses and 42 tulips in vases for table decorations in her restaurant. Each vase will hold the same number of

Bess [88]

the answer is x = 6

6 0

3 years ago

Please do these three and show work

AVprozaik [17]

Answer:

2)16.5

Step-by-step explanation:

step1:v/2.2 -0.1=7.4

add 0.1 on both sides

step2:v/2.2=7.5

multiply by 2.2 on both sides

step3:v=16.5

5 0

3 years ago

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Brandon needs $480 to buy a TV and stereo system for his room. He received $60 in cash for birthday presents. He plans to save $

hammer [34]

The answer is 14 weeks

60 + 30w = 480
30w = 480 - 60
30w = 420
w = 420 / 30
w = 14

3 0

3 years ago

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One of the relatively uncommon instances in which researchers know the population standard deviation is in the case of the Intel

Phoenix [80]

Answer:

90% confidence interval for the mean IQ score in this community is [106.58 , 109.42].

Step-by-step explanation:

We are given that in general, the average IQ score in large, diverse populations is 100 and the standard deviation is 15.

Suppose that your sample of 300 members of your community gives you a mean IQ score of 108.

Firstly, the pivotal quantity for 90% confidence interval for the population mean is given by;

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

where, $\bar X$ = sample mean IQ score = 108

$\sigma$ = population standard deviation = 15

n = sample of members = 300

$\mu$ = population mean IQ score

Here for constructing 90% confidence interval we have used One-sample z test statistics because we know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X -\mu}$ < $1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $108-1.645 \times {\frac{15}{\sqrt{300} } }$ , $108+1.645 \times {\frac{15}{\sqrt{300} } }$ ]

= [106.58 , 109.42]

Therefore, we can be 90% confident that the mean IQ score in this community lies between 106.58 and 109.42.

8 0

3 years ago