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Alex17521 [72]
3 years ago
14

If the cost of 4 pears and 2 peaches equals $1.00 and the cost of 2 pears and 3 peaches equals $0.70 how much does each pear and

each peach cost
Mathematics
2 answers:
soldi70 [24.7K]3 years ago
8 0

Answer:

20 p per pear 10p per peach

Step-by-step explanation:

Hope this helps

zalisa [80]3 years ago
4 0

Answer:

1 pear costs 20 cents. 1 peach costs 10 cents.

Step-by-step explanation:

x represents pears

y represents peaches

4x + 2y = 1

2x + 3y = 0.7

multiply the second equation by -2 to elimate the x variable

(2x +3y = 0.7) -2 = -4x - 6y = -1.4

add the new equation to the first original one:

4x + 2y = 1

<u>-4x - 6y = -1.4</u>

-4y = -0.4. divide both sides by -4, y = 0.10

plug in y = 10, solve algebraically and get x = 0.20

Lastly, check your work:

4(0.20) + 2(0.10) = 1

2(0.20) + 3(0.10) = 0.7

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a) The hypothesis that the mean annual income is the same could not be rejected. There is no enough evidence to claim they are different.

b) The critical values for this two-sided test are t=±1.711.

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>(a) Identify the claim and state  H0  and  Ha. </em>

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We have an hypothesis test on the difference of two means.

The null and alternative hypothesis are:

H_0: \mu_a-\mu_b=0\\\\H_a: \mu_a-\mu_b\neq 0

The claim of the alternative hypothesis is that the mean annual income is different in County A and County B.

The null hypothesis is that the mean annual income is equal in both counties.

The significance level is α=0.10.

The sample from County A has a mean of S40,400, a s.d. of $8,700 and a sample size of 18 residents.

The sample from County B has a mean of S39,200, a s.d. of $6,000 and a sample size of 8 residents.

The standard error of the difference of means is:

\sigma_d=\sqrt{\frac{\sigma_a^2}{n_a}+\frac{\sigma_b^2}{n_b}}=\sqrt{\frac{8700^2}{18}+\frac{6000^2}{8}}=\sqrt{8705000}=2950

The degrees of freedom are:

df=n_a+n_b-2=18+8-2=24

Then, the test statistic is:

t=\frac{\Delta M-\Delta \mu}{\sigma_d} =\frac{(40,400-39,200)-0}{2950} =\frac{1200}{2950}=0.407

For a statistic t=0.407, and df=24, the P-value is P=0.69. As the P-value is bigger than the significance level, the null hypothesis failed to be rejected.

If we would use the critial value approach, we would have to calculate the critical values for t, for df=24, two sided test and α=0.10.

The critical values, looking in a table, are t=1.711.

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