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2 years ago

If a and b vary directly and a = 6 when b = 4, what will be the value of a when b = 16?

Mathematics

1 answer:

SpyIntel [72]2 years ago

7 0

"Vary directly" is another way of saying "directly proportional".

In simple English, this just means, as one increases, the other increases too.

a = 6

b = 4

The question says find out what a is when b is 16.

First, lets find how many times larger b has gotten.

16/4 = 4

b had gotten 4 times bigger to go from b=4 to b=16

If a is varying directly, it means that it will also get 4 times bigger.

Therefore, a = 6 x 4 = 24

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vlada-n [284]

Answer:

Inorder to write a division expression for fraction sasha should divide the numerator (9) by the denominator (14)

Step-by-step explanation:

9√14 which will result 0.643

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3 years ago

Xy+(z-x) x=3 y=2 z=4

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Answer:

Step-by-step explanation:

xy + (z-x)

x = 3

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3 years ago

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3 years ago

Read 2 more answers

How do I work this out?

Maslowich

Here is the work and the answers. Hope this helps! Message for any conversations.

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3 years ago

A college student is interested in investigating the TV-watching habits of her classmates and surveys 20 people on the number of

Alla [95]

Answer:

80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].

Step-by-step explanation:

We are given that a college student is interested in investigating the TV-watching habits of her classmates and surveys 20 people on the number of hours they watch per week. The results are provided below;

Hours of TV per week (X): 6, 14, 13, 6, 16, 10, 19, 4, 5, 5, 18, 8, 7, 14, 8, 8, 9, 12, 6, 5.

Firstly, the Pivotal quantity for 80% confidence interval for the true average is given by;

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

where, $\bar X$ = sample mean number of hours of TV watched per week = $\frac{\sum X}{n}$ = 9.65

s = sample standard deviation = $\sqrt{\frac{\sum (X -\bar X)^{2} }{n-1} }$ = 4.61

n = sample of people = 20

$\mu$ = true average number of hours of TV watched per week

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 true average, $\mu$ is ;

P(-1.33 < $t_1_9$ < 1.33) = 0.80 {As the critical value of t at 19 degrees of

freedom are -1.33 & 1.33 with P = 10%}

P(-1.33 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.33) = 0.80

P( $-1.33 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.33 \times {\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.33 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.33 \times {\frac{s}{\sqrt{n} } }$ ) = 0.80

80% confidence interval for $\mu$ = [ $\bar X-1.33 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.33 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $9.65-1.33 \times {\frac{4.61}{\sqrt{20} } }$ , $9.65+1.33 \times {\frac{4.61}{\sqrt{20} } }$ ]

= [8.28 hours, 11.02 hours]

Therefore, 80% confidence interval of the true average number of hours of TV watched per week is [8.28 hours, 11.02 hours].

7 0

3 years ago