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3 years ago

How are fractions divided the right way?

Mathematics

1 answer:

Lady_Fox [76]3 years ago

8 0

You would keep the first fraction the same , turn the division sign into a multiplication sign , and flip the 2nd fraction upside down.

10/14 ÷ 7/16

10/14 x 16/7= 80/49 = 1 31/49

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PLEASE HELP ASAP!!!!

Fynjy0 [20]

Answer:

It is not valid since there is more than one variable being changed at one.

Step-by-step explanation:

Jackson changed the brand of soil, but he also changes the amount of water, and the amount of sunlight the plants are receiving, so his conclusion that the soil was more effective is not valid.

6 0

3 years ago

Which of these is the algebraic expression for "eight less than some number?"

kipiarov [429]

B-8 would be the answer. This is because we are saying you have 8 less than something. So no matter what something is, we are taking away 8 from it. Since our something is unknown, we use our variable 'b' in its place, and subtract 8, giving us b-8.

8 0

3 years ago

Simplify the expression: -3(6 - 1.5n)=

Tatiana [17]

Answer:

Step-by-step explanation:

Use distributive property: a*(b +c) =a*b + a*c

-3( 6 - 1.5n) = (-3)*6 - (-3)*1.5n

= -18 + 4.5n

3 0

3 years ago

Read 2 more answers

Triangle A'B'C' is formed by a reflection over x = 1 and dilation by a scale factor of 2 from the origin. Which equation shows t

ollegr [7]

Answer:

$\frac{AB}{A"B"}=\frac{1}{2}$

Step-by-step explanation:

Given

$k = 2$ --- scale factor

Required

Relationship between ABC and A"B"C"

$k = 2$ implies that the sides of A"B"C" are bigger than ABC

i.e.

$A"B" = 2AB$

$A"C" = 2AC$

$B"C" = 2BC$

In $A"B" = 2AB$

Divide both sides by A"B"

$1 = \frac{2AB}{A"B"}$

Divide both sides by 2

$\frac{1}{2} = \frac{AB}{A"B"}$

Rewrite as:

$\frac{AB}{A"B"}=\frac{1}{2}$

(a) is correct

8 0

3 years ago

g An irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the

blagie [28]

Answer:

A 90% confidence interval of the true mean is [$119.86, $123.34].

Step-by-step explanation:

We are given that an irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the mean amount of money spent on textbooks was $121.60.

Also, the standard deviation of the population was $6.36.

Firstly, the pivotal quantity for finding the 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 amount of money spent on textbooks = $121.60

$\sigma$ = population standard deviation = $6.36

n = sample of students = 36

$\mu$ = population mean

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

So, 95% 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 $\mu$ = [ $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $121.60-1.645 \times {\frac{6.36}{\sqrt{36} } }$ , $121.60+1.645 \times {\frac{6.36}{\sqrt{36} } }$ ]

= [$119.86, $123.34]

Therefore, a 90% confidence interval of the true mean is [$119.86, $123.34].

5 0

4 years ago