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inn [45]
2 years ago
11

1/4 of the roses are red, 1/3 of the remainder are yellow and the rest are pink. There are 24 more pink roses than red roses. Ho

w many roses are there altogether?
Mathematics
1 answer:
Alika [10]2 years ago
7 0
<h3>There are 96 roses altogether</h3>

<em><u>Solution:</u></em>

Let "x" be the number of roses

From given,

<em><u>1/4 of the roses are red</u></em>

Red\ roses = \frac{1}{4} \times x\\\\Red\ roses = \frac{x}{4}

<em><u>1/3 of the remainder are yellow</u></em>

Remaining = x - \frac{x}{4}\\\\Remaining = \frac{3x}{4}

Therefore,

Yellow\ roses = \frac{1}{3} \times \frac{3x}{4}\\\\Yellow\ roses = \frac{x}{4}

<em><u>Rest are pink</u></em>

Remaining = \frac{3x}{4} - \frac{x}{4}\\\\Remaining = \frac{2x}{4}\\\\Remaining = \frac{x}{2}

Therefore,

Pink\ Roses = \frac{x}{2}

There are 24 more pink roses than red roses

Therefore,

Number of pink roses = 24 + red roses

\frac{x}{2} = 24 + \frac{x}{4}\\\\\frac{x}{2} -  \frac{x}{4} = 24\\\\\frac{x}{4} = 24\\\\x = 24 \times 4\\\\x = 96

Thus there are 96 roses altogether

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Money Spent on Road Repair A politician wishes to compare the variances of the amount of money spent for road repair in two diff
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Answer:

(A) Null Hypothesis, H_0 : \sigma_1^{2} =\sigma_2^{2}      {means that there is no significant difference in the variances of the amounts spent in the two counties}

Alternate Hypothesis, H_A : \sigma_1^{2} \neq \sigma_2^{2}      {means that there is a significant difference in the variances of the amounts spent in the two counties}

(B) The F-table gives critical values of 0.359 and 2.781 for (14,17) degrees of freedom for the two-tailed test.

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(D) We conclude that there is no significant difference in the variances of the amounts spent in the two counties.

Step-by-step explanation:

We are given that a politician wishes to compare the variances of the amount of money spent on road repair in two different counties.

The data are given here below;

County A              County B

s1 = $11,596         s2 = $14,837

n1 = 15                     n2 = 18

Let \sigma_1^{2} = variances of the amounts spent in County A.

\sigma_2^{2} = variances of the amounts spent in County B.

(A) So, Null Hypothesis, H_0 : \sigma_1^{2} =\sigma_2^{2}      {means that there is no significant difference in the variances of the amounts spent in the two counties}

Alternate Hypothesis, H_A : \sigma_1^{2} \neq \sigma_2^{2}      {means that there is a significant difference in the variances of the amounts spent in the two counties}

The test statistics that would be used here <u>Two-sample F-test statistics </u>distribution;

                             T.S. =  \frac{s_1^{2} }{s_2^{2} } \times \frac{\sigma_2^{2} }{\sigma_1^{2} }  ~ F__n_1_-_1,_ n_2_-_1

where, s_1 = sample standard deviation for County A = $11,596

s_2 = sample standard deviation for County B = $14,837

n_1 = sample size for County A = 15

n_2 =  sample size for County B = 18

(B) Now at 0.05 level of significance, the F-table gives critical values of 0.359 and 2.781 for (14,17) degrees of freedom for the two-tailed test.

(C) So, <u><em>the test statistics</em></u>  =  \frac{11,596^{2}  }{14,837^{2} } \times 1  ~  F__1_4,_ 1_7

                                     =  0.611

The value of the F-test statistic is 0.611.

Now, as we can see that our test statistics lie within the range of critical values of F, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

(D) Therefore, we conclude that there is no significant difference in the variances of the amounts spent in the two counties.

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