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SashulF [63]
2 years ago
8

Of the songs in ivan's music library, 4/5 are rock songs. of the rock songs, 3/4 feature a guitar solo. what fraction of the son

gs in ivan's music library are rock songs that feature a guitar solo?
Mathematics
1 answer:
BigorU [14]2 years ago
4 0

In this case this "of" means we will multiply the two fractions "4/5 × 3/4" exists 3/5.

<h3>What are fractions?</h3>

That operation exists utilized to 'simplify' a fraction (reduce it to lower terms).

When there exists no number that goes into both the top and bottom numbers, then the fraction is in simplest form (lowest terms).

We are looking for 4/5 of 3/4. In this case this "of" means we will multiply the two fractions. To multiply fractions, we multiply

top × top and bottom × bottom

4/5 × 3/4 exists 4 × 3 on top and 5 × 4 on the bottom.

= (4 × 3)/(5 × 4)

= 12 / 20

= 3 / 5

Therefore, the correct answer is 3/5.

To learn more about fractions refer to:

brainly.com/question/571030

#SPJ4

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Step-by-step explanation:

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How do you find the answer to this?
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Answer:

  D)  {-1}

Step-by-step explanation:

A rational expression is "not defined" when its denominator is zero. Hence to find values of x that make the expression "not defined," you solve the equation ...

  denominator = 0

  2x +2 = 0 . . . . . . put the actual denominator into the equation

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3 years ago
How do you turn a fahrenheit to kelvins??
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6 0
3 years ago
A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime
MrRissso [65]

Answer:

We conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

We conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

Step-by-step explanation:

We are given a random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen;

1.10, 5.09, 0.97, 1.59, 4.60, 0.32, 0.55, 1.45, 0.14, 4.47, 1.20, 3.50, 5.02, 4.67, 5.22, 2.69, 3.98, 3.17, 3.03, 2.21, 0.69, 4.47, 3.31, 1.17, 0.76, 1.17, 1.57, 2.62, 1.66, 2.05.

Let \mu = <u><em>true average percentage of organic matter</em></u>

So, Null Hypothesis, H_0 : \mu = 3%      {means that the true average percentage of organic matter in such soil is 3%}

Alternate Hypothesis, H_A : \mu \neq 3%      {means that the true average percentage of organic matter in such soil is something other than 3%}

The test statistics that will be used here is <u>One-sample t-test statistics</u> because we don't know about the population standard deviation;

                         T.S.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean percentage of organic matter = 2.481%

             s = sample standard deviation = 1.616%

            n = sample of soil specimens = 30

So, <u><em>the test statistics</em></u> =  \frac{2.481-3}{\frac{1.616}{\sqrt{30} } }  ~ t_2_9

                                     =  -1.76

The value of t-test statistics is -1.76.

(a) Now, at 10% level of significance the t table gives a critical value of -1.699 and 1.699 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics doesn't lie within the range of critical values of t, so we have <u><em>sufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

(b) Now, at 5% level of significance the t table gives a critical value of -2.045 and 2.045 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have <u><em>insufficient evidence to reject our null hypothesis</em></u> as it will not fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

8 0
3 years ago
Ellen saved $2000 and she spent 10% of it to buy a bike. Kim saved 8% more than that of Ellen's left savings. If Lisa saved 60%
enyata [817]

Answer: Lis saved $1166.4 .

Step-by-step explanation:

Given , Ellen saved $2000 and she spent 10% of it to buy a bike.

Then, Ellen's left saving = saved amount - 10% of  (saved amount)

= $2000 - 0.10 x $2000 <em>[We convert perecntage to decimal by diving it by 100]</em>

= $2000 - $200

= $1800

So, Ellen's left saving= $1800

Also, Kim saved 8% more than that of Ellen's left savings.

Kim's Saving = Ellen's left saving + 8% of (Ellen's left saving)

= $1800  + (0.08) x ($1800) <em>[We convert perecntage to decimal by diving it by 100]</em>

= $1800 +  $144

i.e. Kim's Saving= $1944

If Lisa saved 60% as much as Kim, then Lisa's saving  = 60% of (Kim's Saving)

= (0.60) x ($1944) <em>[We convert perecntage to decimal by diving it by 100]</em>

i.e. Lisa's saving = $1166.4

4 0
3 years ago
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