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s2008m [1.1K]
3 years ago
15

#40 please! Thank you.

Mathematics
1 answer:
lisov135 [29]3 years ago
5 0
Well he and the cart together weigh: 210+95=305.
The weight left for boxes is: 4,000-305=3,695.
Total weight of boxes is: 50·275=13,750.
Divide the weight of boxes by how much is left for the elevator: 13,750/3,695=3.72
Since 3.72 is more than 3 he would have to make 4 trips to safely do so :)
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If Sarah turns 15 on august 27 and her graduation year is 2024 how old will she be when she graduates high school?
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he owner of a local supermarket wants to estimate the difference between the average number of gallons of milk sold per day on w
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Answer:

90% confidence interval is ( -149.114, -62.666   )

Step-by-step explanation:

Given the data in the question;

Sample 1                                Sample 2

x"₁ = 259.23                            x"₂ = 365.12

s₁  = 34.713                              s₂ = 48.297

n₁ = 5                                       n₂ = 10

With 90% confidence interval for μ₁ - μ₂ { using equal variance assumption }

significance level ∝ = 1 - 90% = 1 - 0.90 = 0.1

Since we are to assume that variance are equal and they are know, we will use pooled variance;

Degree of freedom DF = n₁ + n₂ - 2 = 5 + 10 - 2 = 13

Now, pooled estimate of variance will be;

S_p^2 = [ ( n₁ - 1 )s₁² + ( n₂ - 1)s₂² ] / [ ( n₁ - 1 ) + ( n₂ - 1 ) ]

we substitute

S_p^2 = [ ( 5 - 1 )(34.713)² + ( 10 - 1)(48.297)² ] / [ ( 5 - 1 ) + ( 10 - 1 ) ]

S_p^2 = [ ( 4 × 1204.9923) + ( 9 × 2332.6 ) ] / [  4 + 9 ]

S_p^2 = [ 4819.9692 + 20993.4 ] / [  13 ]

S_p^2 = 25813.3692 / 13

S_p^2 = 1985.64378

Now the Standard Error will be;

S_{x1-x2 = √[ ( S_p^2 / n₁ ) + ( S_p^2 / n₂ ) ]

we substitute

S_{x1-x2 = √[ ( 1985.64378 / 5 ) + ( 1985.64378 / 10 ) ]

S_{x1-x2 = √[ 397.128756 + 198.564378 ]

S_{x1-x2 = √595.693134

S_{x1-x2 = 24.4068

Critical Value = t_{\frac{\alpha }{2}, df = t_{0.05, df=13 = 1.771  { t-table }

So,

Margin of Error E =  t_{\frac{\alpha }{2}, df × [ ( S_p^2 / n₁ ) + ( S_p^2 / n₂ ) ]

we substitute

Margin of Error E = 1.771 × 24.4068

Margin of Error E = 43.224

Point Estimate = x₁ - x₂ = 259.23 - 365.12 = -105.89

So, Limits of 90% CI will be; x₁ - x₂ ± E

Lower Limit = x₁ - x₂ - E = -105.89 - 43.224 = -149.114

Upper Limit = x₁ - x₂ - E = -105.89 + 43.224 = -62.666

Therefore, 90% confidence interval is ( -149.114, -62.666   )

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