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3 years ago

For 7/20 what is an equivalent fraction with a denominator of 100

Mathematics

2 answers:

stiv31 [10]3 years ago

7 0

100/20=5
7×5=35
35/100

Paladinen [302]3 years ago

5 0

Start with 7/20 .

We can multiply the denominator by anything we want to,
but if we do that, we must immediately multiply the numerator
by the same thing, to avoid changing the value of the fraction.
(I guess you could call this process "un-simplifying" the fraction.)

The denominator is 20 right now, and we want it to be 100 .
So we have to multiply it by 5 .

Immediately, before we forget, we must also multiply the numerator by 5 .

So we end up with 35/100 .

This fraction has exactly the same value as 7/20 .
You should prove that for yourself by simplifying 35/100 .

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Kevin can mow the lawn in 1.5 hours. Together, Kevin and Eric can mow the lawn in 30 minutes. How long will it take Eric to mow

xeze [42]

Answer:

Step-by-step explanation:

These work problems are always done in terms of fractions ie if both (Kevin & Eric) took 30mins to mow and Kevin took 1.5hrs (90mins) alone then we can make an equation like below

1/time took for both = 1/time took for Kevin + 1/time took for Eric

1/30 = 1/90+1/x ==> calling time took for Kevin x

solving the above 1/x = 1/30-1/90

1/x=2/90

x=45

6 0

3 years ago

What a impor fraction

MrMuchimi

Im guessing you meant improper. This means that the numerator is greater than the denominator. An example is 11/4

4 0

3 years ago

Read 2 more answers

An automobile company wants to determine the average amount of time it takes a machine to assemble a car. A sample of 40 times y

aksik [14]

Answer:

A 98% confidence interval for the mean assembly time is [21.34, 26.49] .

Step-by-step explanation:

We are given that a sample of 40 times yielded an average time of 23.92 minutes, with a sample standard deviation of 6.72 minutes.

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 average time = 23.92 minutes

s = sample standard deviation = 6.72 minutes

n = sample of times = 40

$\mu$ = population mean assembly time

Here for constructing a 98% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, a 98% confidence interval for the population mean, $\mu$ is;

P(-2.426 < $t_3_9$ < 2.426) = 0.98 {As the critical value of z at 1% level

of significance are -2.426 & 2.426}

P(-2.426 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.426) = 0.98

P( $-2.426 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.426 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

P( $\bar X-2.426 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.426 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X-2.426 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.426 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $23.92-2.426 \times {\frac{6.72}{\sqrt{40} } }$ , $23.92+2.426 \times {\frac{6.72}{\sqrt{40} } }$ ]

= [21.34, 26.49]

Therefore, a 98% confidence interval for the mean assembly time is [21.34, 26.49] .

7 0

3 years ago

5x - (x + 3)= 1/3 (9x + 18) - 5

telo118 [61]

1/3(9x) =3x
1/3(18) =6

3x+6-5

Then you do
-(x)=-x
-(3)=-3

5x-x-3 = 4x-3

4x-3=3x+6-5
4x-3=3x+1

x=4

4 0

3 years ago

Read 2 more answers

An NFL coach sometimes uses a defense that utilizes 3 defensive linemen, 4 linebackers, and 4 defensive backs. His roster (the p

Alexandra [31]

Answer:

68,600

Step-by-step explanation:

The order of the players is not important. For example, a defensive line of Shaq Lawson, Ed Oliver and Jerry Hughes is the same as a defensive line of Ed Oliver, Shaq Lawson and Jerry Hughes. So we use the combinations formula to solve this question.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

Defensive Lineman:

3 from a set of 8. So

$C_{8,3} = \frac{8!}{3!5!} = 56$

56 combinations of defensive lineman

Linebackers:

4 from a set of 7. So

$C_{7,4} = \frac{7!}{4!3!} = 35$

35 combinations of linebackers

Defensive backs:

4 from a set of 7. So

$C_{7,4} = \frac{7!}{4!3!} = 35$

35 combinations of defensive backs

How many different ways can the coach pick the 11 players to implement this particular defense?

56*35*35 = 68,600

68,600 different ways can the coach pick the 11 players to implement this particular defense

7 0

3 years ago