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2 years ago

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I WILL GIVE BRAINLIEST The fuel efficiency of Natalie's car is 25 miles per gallon. Which of the following statements are true?

Select two that apply.
a. For every mile Natalie travels, she uses 25 gallons.
b. For every gallon Natalie uses, she travels 0.04 miles.
c. For every mile Natalie travels, she uses 0.04 gallons.
d. For every gallon Natalie uses, she travels 25 miles.

1 answer:

sleet_krkn [62]2 years ago

8 0

Answer:

The fuel efficiency of Natalie's car is 25 miles per gallon. Which of the following statements are true? Select two that apply.

a. For every mile Natalie travels, she uses 25 gallons.

b. For every gallon Natalie uses, she travels 0.04 miles.

✓c. <u>For every mile Natalie travels, she uses 0.04 gallons.</u>

✓d. <u>For every gallon Natalie uses, she travels 25 miles.</u>

C. and D. are the right answer.

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Geometry homework! help

Bogdan [553]

Answer:

cos∠ABC = 12/15 = 0.80

tan∠ABC = 9/12 = 0.75

sin∠ACB = 12/15 = 0.80

cos∠ACB = 9/15 = 0.60

tan∠ACB = 12/9 = 1 1/3

Step-by-step explanation:

Find the Adjacent, Opposite, or Hypotenuse angles to the angles in the picture and fill them into the given fraction.

cos∠ABC = Adjacent/Hypotenuse = 12/15 = 4/5 = 0.80

tan∠ABC = Opposite/Adjacent = 9/12 = 3/4 = 0.75

sin∠ACB = Opposite/Hypotenuse = 12/15 = 4/5 = 0.80

cos∠ACB = Adjacent/Hypotenuse = 9/15 = 3/5 = 0.60

tan∠ACB = Opposite/Adjacent = 12/9 = 1 1/3 (1 1/3 in decimal form is 1.3 with a line on top of 3 (Bar on top of 3))

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2 years ago

Match each expression with an equivalent expression.

Marysya12 [62]

Answer:

$6x-4y$ → 4. $2(3x-2y)$

$6-4x$ → 3. $2(3-2x)$

$6x-4x$ → 1. $2x$

$6x-4$ → 2. $2(3x-2)$

Step-by-step explanation:

You need to:

Factor out 2 from $6x-4y$ , then you get the equivalent expression:

$=2(3x-2y)$ this matches with OPTION 4

Factor out 2 from $6-4y$ , then you get the equivalent expression:

$=2(3-2x)$ this matches with OPTION 3

Make the subtraction for $6x-4x$ , then you get the equivalent expression:

$=2x$ this matches with OPTION 1

Factor out 2 from $6x-4$ , then you get the equivalent expression:

$=2(3x-2)$ this matches with OPTION 2

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2 years ago

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Pls answer me this all question

Scorpion4ik [409]

Answer:

a. $5 \frac{21}{40}$

b. $\frac{1}{42}$

Step-by-step explanation:

$a. \: 12 \frac{2}{5} - 6 \frac{7}{8}$

$\frac{62}{5} - \frac{55}{8}$

$\frac{62}{8} \times \frac{8}{8} - \frac{55}{8} \times \frac{5}{5}$

$\frac{62 \times 8 - 55 \times 5}{40}$

$\frac{221}{40}$

$5 \frac{21}{40}$

$b. \: \frac{3}{14} - \frac{4}{21}$

$\frac{3}{14} \times \frac{3}{3} - \frac{4}{21} \times \frac{2}{2}$

$\frac{3 \times 3}{42} - \frac{4 \times 2}{42}$

$\frac{1}{42}$

Hope it is helpful....

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2 years ago

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Day care centers expose children to a wider variety of germs than the children would be exposed to if they stayed at home more o

Ostrovityanka [42]

Answer:

a) The study is an observational study

b) The proportion of children with significant social activity in children with acute lymphoblastic leukemia is approximately 0.802

The proportion of children with significant social activity in children without acute lymphoblastic leukemia is approximately 0.8565

c) The odds ratio is approximately 0.6780

d) The 95% CI is approximately 0.523 < OR < 0.833

e) i) Yes

ii) Children with more social activity have a higher occurrence of acute lymphoblastic leukemia

Step-by-step explanation:

a) An experimental study is a study in which the researcher adds inputs to the study and then monitors the outcomes

An observational study is one in which the researcher observes risk factors and does not intervene in the process

Therefore, the study is an observational study

b) The proportion of children with significant social activity in children with acute lymphoblastic leukemia is $\hat p_1$ = 1020/1272 = 85/106 = $0.8\overline {0188679245283}$ ≈ 0.802

The proportion of children with significant social activity in children without acute lymphoblastic leukemia is $\hat p_2$ = 5343/6238 ≈ 0.8565

c) We have the following two way table;

$\begin{array}{ccc}Lymphoblastic \ Leukemia &With \ Social \ Activity & Without \ Social \ Activity\\With&1020 \ (a)&252 \ (b)\\Without&5343 \ (c)&895 \ (d) \end{array}$

$The \ odds \ ratio = \dfrac{1,020 \times 895}{5,343 \times 252} \approx 0.6780$

The odds ratio ≈ 0.6780

d) The 95% confidence interval for the odds ratio is given as follows;

$95 \% \ CI = OR \ \pm \ 1.96 \times \sqrt{\dfrac{1}{a} +\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}}$

Where;

a = 1020, b = 252, c = 5343, d = 895

Therefore;

$95 \% \ CI \approx 0.6780 \ \pm \ 1.96 \times \sqrt{\dfrac{1}{1,020} +\dfrac{1}{252}+\dfrac{1}{5,343}+\dfrac{1}{895}} \approx 0.678 \pm0.155$

The 95% CI ≈ 0.523 < OR < 0.833

e) i) Given that the observed Odds Ratio is within the Confidence Interval, therefore, there is an indication that the amount of social activity is associated with acute lymphoblastic leukemia

ii) Based on the proportion of the study findings, children with more social activity have a higher occurrence of acute lymphoblastic leukemia

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2 years ago

Calculate the future value of $7,000 earning 6% interest compounded quarterly for 7 years. (Round your answer to two decimal pla

qwelly [4]

Use the compound interest formula.

Let A = the ending amount

Let P = the principal

Let r = the interest rate

Let n = the amount compounded a year

Let t = time

A = P(1 + r/n) ^(n/t)

Substitute your numbers in

A = $7,000(1 + 0.06/4)^(4/7)

Solve for A

A = $7,059.81

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3 years ago