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2 years ago

I need help please

Mathematics

2 answers:

Sidana [21]2 years ago

7 0

7 1/2
you go to both sides of the graph and count in till u get to the middle

Marat540 [252]2 years ago

3 0

Answer:

7 1/2

Step-by-step explanation:

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Bryan has a balance of 320 in his checking account he spends 15 each week for the next eight weeks. write an equation showing hi

viva [34]

Bryan spends T=15x
320 divided by 15 is and average of $21 per week

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2 years ago

Nico earns $32.50 for each hour he works. Last week Nico earned $1,300, and this week he earned $1,040. How many more hours did

ladessa [460]

Answer:

8 hours

Step-by-step explanation:

$1,300/$32.50 = 40

$1,040/$32.50 = 32

40-32=8

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2 years ago

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A study was conducted to determine if there was a difference in the driving ability of students from West University and East Un

Sholpan [36]

Answer:

There is no significant evidence which shows that there is a difference in the driving ability of students from West University and East University, assuming a significance level 0.1

Step-by-step explanation:

Let p1 be the proportion of West University students who involved in a car accident within the past year

Let p2 be the proportion of East University students who involved in a car accident within the past year

Then

$H_{0}:$ p1=p2

$H_{a}:$ p1≠p2

The formula for the test statistic is given as:

z= $\frac{p1-p2}{\sqrt{\frac{p*(1-p)*(n1+n2)}{n1*n2} } }$ where

p1 is the sample proportion of West University students who involved in a car accident within the past year (0.15)
p2 is the sample proportion of East University students who involved in a car accident within the past year (0.12)

p is the pool proportion of p1 and p2 ( $\frac{15+12}{100+100}=0.135$ )
n1 is the sample size of the students from West University (100)
n2 is the sample size ofthe students from East University (100)

Then we have z= $\frac{0.03}{\sqrt{\frac{0.135*0.865*(100+100)}{100*100} } }$ ≈ 0.6208

Since this is a two tailed test, corresponding p-value for the test statistic is ≈ 0.5347.

Assuming significance level 0.1, The result is not significant since 0.5347>0.1. Therefore we fail to reject the null hypothesis at 0.1 significance

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2 years ago

Two processes are used to produce forgings used in an aircraft wing assembly. Of 200 forgings selected from process 1, 10 do not

tangare [24]

Answer:

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

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:-

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

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:

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