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Fynjy0 [20]
1 year ago
5

Laura will go for a run during her lunch break if the temperature is between 60∘F

Mathematics
1 answer:
balu736 [363]1 year ago
8 0

Answer:

D(t) = 3 + 0.0(80 - t)

Step-by-step explanation:

The average of speed of Laura in miles per hour is given by:

S(t) = 6 + 0.1(80 - t)                                 Equation 1

where, t is the temperature in degrees Fahrenheit.

The distance D, Laura covers at x miles per hour is given as:

D(x) = 0.5x                                              Equation 2

We need to find the expression that models the distance that Laura runs in terms of the temperature "t"

The "x" in Equation 1 represents the average speed of Laura in miles per hour. S(t) in Equation 1 also represent the speed of Laura in miles per hour. So, we can replace x by S(t) in Equation 2 and generate an equation of Distance in terms of temperature "t" as shown below:

D(S(t)) = 0.5 (6 + 0.1(80-t))

D(t) = 3 + 0.0(80 - t)

This expression models the distance that Laura runs in 30 minutes given that it is t∘F outside at the start of her run.

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Large flower pots weigh 28 pounds each.
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Two processes are used to produce forgings used in an aircraft wing assembly. Of 200 forgings selected from process 1, 10 do not
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Answer:

a.

\bar p_1=0.05\\\bar p_2=0.067

b-Check illustration  below

c.(-0.0517,0.0177

Step-by-step explanation:

a.let p_1  \& p_2 denote processes 1 & 2.

For p_1: T1=10,n1=200

For p_2:T2=20,n2=300

Therefore

\bar p_1=\frac{t_1}{N_1}=\frac{10}{200}=0.05\\\bar p_2=\frac{t_2}{N_2}=\frac{20}{300}=0.067

b. To test for hypothesis:-

i.

H_0:p_1=p_2\\H_A=p_1\neq p_2\\\alpha=0.05

ii.For a two sample Proportion test

Z=\frac{\bar p_1-\bar p_2}{\sqrt(\bar p(1-\bar p)(\frac{1}{n_1}+\frac{1}{n_2})}\\

iii. for \frac{\alpha}{2}=(-1.96,+1.96) (0.5 alpha IS 0.025),

reject H_o if|Z|>1.96

iv. Do not reject H_o. The noncomforting proportions are not significantly different as calculated below:

z=\frac{0.050-0.067}{\sqrt {(0.06\times0.94)\times \frac{1}{500}}}

z=-0.78

c.(1-\alpha).100\% for the p1-p2 is given as:

(\bar p_1-\bar p_2)\pm Z_0_._5_\alpha \times \sqrt   \frac{ \bar p_1(1-\bar p_1)}{n_1}+\frac{\bar p_2(1-\bar p_2)}{n_2}\\\\=(0.05-0.067)\pm 1.645  \times \sqrt \ \frac{0.05+0.95}{200}+\frac{0.067+0.933}{300}\\

=(-0.0517,+0.0177)

*CI contains o, which implies that proportions are NOT significantly different.

4 0
2 years ago
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