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3 years ago

Ms. Bedru conducts a survey of the ages

Mathematics

1 answer:

lorasvet [3.4K]3 years ago

5 0

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You might be interested in

The cost of renting a car is $85/wk plus $0.15/mi traveled during that week. An equation to represent the cost would be y = 85 +

meriva

Answer:

I was being charged for 65 miles

Step-by-step explanation:

Given

$y = 85 + 0.15x$

Required

Find x when y = 94.75

Substitute 94.75 for y in $y = 85 + 0.15x$

$94.75 = 85 + 0.15x$

Collect Like Terms

$0.15x = 94.75 - 85$

$0.15x = 9.75$

Solve for x

$x = \frac{9.75}{0.15}$

$x = 65\\$

I was being charged for 65 miles

6 0

4 years ago

What are the coordinates of the midpoint of the line segment with endpoints R(4, -7) and S(-3, 5)?

bezimeni [28]

Answer:

(1/2, -1)

Step-by-step explanation:

The midpoint is the average of the end points:

M = (R + S)/2 = ((4, -7) + (-3, 5))/2 = ((4-3)/2, (-7+5)/2) = (1/2, -2/2)

M = (1/2, -1)

5 0

4 years ago

How to find the rule for the nth term of a geometric sequence with changing common ratio?

ki77a [65]

Hmmm... a geometric sequence MUST have a fixed common ratio. If it is changing, then the sequence you are looking at might not be a geometric sequence at all. We'd need to see an example to be sure.

6 0

3 years ago

for the data values below construct a 95 confidence interval if the sample mean is known to be 12898 and the standard deviation

Volgvan

Answer:

A 95% confidence interval for the population mean is [3315.13, 22480.87] .

Step-by-step explanation:

We are given that for quality control purposes, we collect a sample of 200 items and find 24 defective items.

Firstly, the pivotal quantity for finding the 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 proportion of defective items = 12,898

s = sample standard deviation = 7,719

n = sample size = 5

$\mu$ = population mean

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

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

P(-2.776 < $t_4$ < 2.776) = 0.95 {As the critical value of t at 4 degrees of

freedom are -2.776 & 2.776 with P = 2.5%}

P(-2.776 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.776) = 0.95

P( $-2.776 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.776 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.776 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.776 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.776 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.776 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $12,898-2.776 \times {\frac{7,719}{\sqrt{5} } }$ , $12,898+2.776 \times {\frac{7,719}{\sqrt{5} } }$ ]

= [3315.13, 22480.87]

Therefore, a 95% confidence interval for the population mean is [3315.13, 22480.87] .

5 0

3 years ago

Answer b,c,a,b please. I need It for school and I can’t figure it out reach out ASAP

Verdich [7]

Answer:b

Step-by-step explanation:the height for :B: is 14 units long 6/3

7 0

3 years ago