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meriva
2 years ago
10

You work at a fruit market. Bananas cost 50¢ a pound. A customer hands you a bunch of bananas that weighs 3 pounds. How much sho

uld you charge for the bunch of bananas?
Mathematics
2 answers:
weqwewe [10]2 years ago
5 0
3 pounds of bananas cost $1.50
Goshia [24]2 years ago
4 0

Answer:  I should charge 150¢ for the bunch of bananas.  

Step-by-step explanation:  Given that I work at a fruit market and bananas cost 50¢ a pound.

A customer hands me a bunch of bananas that weighs 3 pounds.

We are to find the price that I should charge for the bunch of bananas.

We will be using the UNITARY method to solve the given problem.

Also, we know that

100 ¢ = $1.

Cost of 1 pound of bananas =  50¢

Therefore, the cost of 3 pounds of bananas is given by

p = (50 × 3)¢ = 150¢ = $1.50.

Thus, I should charge 150¢ for the bunch of bananas.  

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Mustafa’s soccer team is planning a school dance as a fundraiser. The DJ charges $200 and decorations cost $100. The team decide
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The success of an airline depends heavily on its ability to provide a pleasant customer experience. One dimension of customer se
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A)H0: μ1 − μ2 = 0

Ha: μ1 − μ2 ≠ 0

(b)  Means

Company A___50.6___ min.

Company B___52.75___ min.

c)The t-value is -0.30107.

The p-value is 0 .764815.

C) Do not reject H0. There is no statistical evidence that one airline does better than the other in terms of their population mean delay time.

Step-by-step explanation:

A)H0: μ1 − μ2 = 0

i.e there is no difference between the means of delayed flight for two different airlines

Ha: μ1 − μ2 ≠ 0

i.e there is a difference between the means of delayed flight for two different airlines

(b)

Company A___50.6___ min.

Company B___52.75___ min.

Mean of Company A = x`1= ∑x/n =34+ 59+ 43+ 30+ 3+ 32+ 42+ 85+ 30+ 48+ 110+ 50+ 10+ 26+ 70+ 52+ 83+ 78+ 27+ 70+ 27+ 90+ 38+ 52+ 76/25

= 1265/25= 50.6

Mean of Company B = x`2= ∑x/n =

=46+ 63+ 43+ 33+ 65+ 104+ 45+ 27+ 39+ 84+ 75+ 44+ 34+ 51+ 63+ 42+ 34+ 34+ 65+ 64/20

= 1055/20= 52.75

<u><em>Difference Scores Calculations</em></u>

Company A

Sample size for Company A= n1= 25

Degrees of freedom for company A= df1 = n1 - 1 = 25 - 1 = 24

Mean for Company A= x`1=  50.6

Total Squared Difference (x-x`1) for Company A= SS1: 16938

s21 = SS1/(n1 - 1) = 16938/(25-1) = 705.75

Company B

Sample size for Company B= n1= 20

Degrees of freedom for company B= df2 = n2 - 1 = 20 - 1 = 19

Mean for Company B= x`2=  52.75

Total Squared Difference (x-x`2) for Company B= SS2=7427.75

s22 = SS2/(n2 - 1) = 7427.75/(20-1) = 390.93

<u><em>T-value Calculation</em></u>

<u><em>Pooled Variance= Sp²</em></u>

Sp² = ((df1/(df1 + df2)) * s21) + ((df2/(df2 + df2)) * s22)

Sp²= ((24/43) * 705.75) + ((19/43) * 390.93) = 566.65

s2x`1 = s2p/n1 = 566.65/25 = 22.67

s2x`2 = s2p/n2 = 566.65/20 = 28.33

t = (x`1 - x`2)/√(s2x`1 + s2x`2) = -2.15/√51 = -0.3

The t-value is -0.30107.

The total degrees of freedom is = n1+n2- 2= 25+20-2=43

The critical region for two tailed test at significance level ∝ =0.05 is

t(0.025) (43) = t > ±2.017

Since the calculated value of t=  -0.30107.  does not fall in the critical region t > ±2.017, null hypothesis is not rejected that is there is no difference between the means of delayed flight for two different airlines.

The p-value is 0 .764815. The result is not significant at p < 0.05.

7 0
1 year ago
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