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2 years ago

Hey i needed help with this scale factor

Mathematics

1 answer:

WITCHER [35]2 years ago

7 0

X=16
Divide 44 by 11 you get 4
Same with 52 and 13
Multiply 4 and 4

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Divide 21a^3-14a by -7 A.-3a^3+2a B.-3a^3-2a C.-3a^2+2 D.3a^3+2a

Juli2301 [7.4K]

Answer:

The correct answer is A:-3a^3+2a

4 0

3 years ago

Read 2 more answers

Using the transformation T: (x, y) (x + 2, y + 1), find the distance named. Find the distance BB'

Shalnov [3]

I think its √5, but I'm not entirely sure.

7 0

3 years ago

What is the sum of the polynomials?

LenaWriter [7]

Answer:

6x+4+3x²

Step-by-step explanation:

Given polynomials:

(6x+7+x²)+(2x²-3)

Solution:

(6x+7+x²)+(2x²-3)

Removing brackets,

6x+7+x²+2x²-3

Collecting like terms,

6x+7-3+x²+2x²

Simplifying ,

6x+4+3x²

Done!

4 0

2 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

The amount of time Jill trained for a marathon this month is 125% of the amount of time she trained last month. Jill trained fo

Liono4ka [1.6K]

Answer:

Step-by-step explenation . First you know that the amount of time she trained for the marathon is 125% of the time she trained last month. We also know that she trained 80 hours last month. And the question wants us to know how many hours she trained this month. So we do 100% =x

And 125% = 80 so we cross multiply it becomes 100 X 80 =125x witch eaquals to 8,000 divided by 125 witch is 64. So the final Answer is 64 . IF I AM CORRECT PLEASE MARK BRAINLIEST.

7 0

3 years ago