Ask question

Ask question

All categories

1 year ago

8

The average price for gasoline per gallon is $3.562 if last year the average price for gasoline per gallon was $2.810 what is th

e approximate percent of change

1 answer:

Greeley [361]1 year ago

4 0

Answer: It increased by 78.89%

Step-by-step explanation:

2.810 / 3.562

=0.7889

78.89%

You might be interested in

The table gives the probabilities that orphaned pets in animal shelters in six cities are one of the types listed.

vagabundo [1.1K]

Answer:

the ablity to choose one is the right now

Step-by-step explanation:

3 0

1 year ago

To make lemonade, you used one quart of water, one pint of lemon juice, and 9 ounces of your secret ingredient: honey. How much

WITCHER [35]

To solve this problem you must follow the proccedure shown below:

1. Amount of lemonade in pints:

1 <span>quart of water=2 pints
1 pint of lemon
(9 ounces of honey)(0.0625 pints/1 ounce)=0.56 pints

Total in pints=2 pints+1 pint+0.56 pints
Total in pints=3.56 pints

2. </span>Amount of lemonade in<span> cups:

Total in cups=(3.56 pints)(2 cups/1 pint)
Total in cups=7.12 cups

3. </span>Amount of lemonade in<span> ounces:

Total in ounces=(7.12 cups)(8 ounces/1 cup)
Total in ounces=56.96 ounces</span>

4 0

1 year ago

What error did Leah make?

Evgen [1.6K]

Answer:

the lines arent parallel, so you cant use corresponding angles theorem

3 0

1 year 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:

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

John Johnson is forty-seven. He is purchasing twenty-year endowment insurance, with a face value of $20,000. What is his annual

Alenkinab [10]

The answer is Letter D - 939.80.

You can refer to the attachment for the rate. Since he is forty-seven years old, use that age to find his rate under a twenty-year endowment insurance. In this case, the rate is 46.99. Multiply that rate to 20 since he purchased a 20-year endowment insurance with a face value of $20,000. (20,000/1,000 = 20)

46.99 x 20 = 939.80

4 0

2 years ago