Nora is 4 years older than Hina . If the sum of their ages is at most 45 years ,find the maximum possible age of Hina. *
1 answer:
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cost of 1 rose = $ 3
cost of 1 carnation = $ 1
cost of 1 tulip = $ 3
<em><u>Solution:</u></em>
Let "r" be the cost of 1 rose
Let "c" be the cost of 1 carnation
Let "t" be the cost of 1 tulip
<em><u>3 roses, 2 carnations, and 1 tulip cost $14</u></em>
So we can frame a equation as:
3 roses x cost of 1 rose + 2 carnations x cost of 1 carnation + 1 tulip x cost of 1 tulip = $ 14
3r + 2c + 1t = 14 ----- eqn 1
<em><u>6 roses, 2 carnations, and 6 tulips cost $38</u></em>
So we can frame a equation as:
6 roses x cost of 1 rose + 2 carnations x cost of 1 carnation + 6 tulip x cost of 1 tulip = $ 38
6r + 2c + 6t = 38 ------ eqn 2
<u><em>1 rose, 12 carnations, and 1 tulip cost $18</em></u>
So we can frame a equation as:
1 rose x cost of 1 rose + 12 carnations x cost of 1 carnation + 1 tulip x cost of 1 tulip = $ 18
r + 12c + t = 18 ----- eqn 3
<em><u>Let us solve eqn 1 and eqn 2 and eqn 3 to find values of "r" "c" "t"</u></em>
3r + 2c + 1t = 14 ----- eqn 1
6r + 2c + 6t = 38 ------ eqn 2
r + 12c + t = 18 ----- eqn 3
From eqn 1,
3r = 14 - 2c - t
Substitute the above value of r in eqn 2
Substitute c = 2t - 5 and in eqn 3
Substitute t = 3 in c = 2t - 5
c = 2(3) - 5
<h3>c = 1</h3>Substitute t = 3 and c = 1 in
<em><u>Summarizing the results:</u></em>
cost of 1 rose = $ 3
cost of 1 carnation = $ 1
cost of 1 tulip = $ 3
Answer:
a. Mean
b. μ = μ₀
c. 11.5%
d. Yes
Step-by-step explanation:
The given parameters are;
The mean work week for engineers, μ₀ = 63 hours
The standard deviation, σ = 5 hours
The number of engineers in Kara's sample, n = 10 engineers
The responses given by the 10 engineers are;
70; 45; 55; 60; 65; 55; 55; 60; 50; 55
a. The given information which the newly hired is hoping to find out that it is not is the <em>mean</em>
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b. The null hypothesis which is that the company claim is correct, is therefore;
Null hypothesis, H₀; μ = μ₀ = 63 hours
c. Kara's mean, is found as follows;
= (70 + 45 + 55 + 60 + 65 + 55 + 55 + 60 + 50 + 55)/10 = 57
= 57 hours
The Z-score is therefore;
Z = (57 - 63)/5 = -1.2
From the Z-table, we have;
The p-value for P( ≤ 57) = P(Z ≤ -1.2) = 0.11507
The probability as a percentage given to one decimal place, that the mean would be as low as Kara's mean, P( ≤ 57) = 11.5%
d. Given that the percent percentage chance (11.5%) that the mean could be as low as hers ( = 57 hours) is higher than 10%, she should accept the claim.