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3 years ago

14

Jake took $20.00 to the shop and asked that it be changed into 5-cent coins. How many coins did he get?

1 answer:

coldgirl [10]3 years ago

8 0

Answer:

He got 400 coins because there are 100 cents in a dollar. 100*20=2000. 2000/5=400!

Step-by-step explanation:

I hope this helps! Have a great rest of your day!

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If f(x)=x-5 and g(x)=3x+11, find (f+g)(x).

lawyer [7]

Answer:

4x+6

Step-by-step explanation:

x-5+(3x+11)=4x+6

4 0

3 years ago

Can someone help me solve question 2 and 3?

ale4655 [162]

Answer:

Question 2 = 15 drinks

Question 3 = 7/12

Step-by-step explanation:

Question 2:

For every drink, there are three sizes

Ratio = 1:3

There are 5 flavors

Multiply each side by 5

5:15

15 drinks

Question 3:

5 red, 6 yellow, 8 blue, 3 orange, 2 purple

Possible outcomes = 5 + 6 + 8 + 3 + 2 = 24 possible outcomes

Favourable outcomes = 6 + 8 = 14 favourable outcomes

Fraction = 14/24 = 7/12

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

8 0

3 years ago

The author drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 199 times. Here are the observed

Anton [14]

Answer with explanation:

An Unbiased Dice is Rolled 199 times.

Frequency of outcomes 1,2,3,4,5,6 are=28, 29, 47, 40, 22, 33.

Probability of an Event

$=\frac{\text{Total favorable Outcome}}{\text{Total Possible Outcome}}\\\\P(1)=\frac{28}{199}\\\\P(2)=\frac{29}{199}\\\\P(3)=\frac{47}{199}\\\\P(4)=\frac{40}{199}\\\\P(5)=\frac{22}{199}\\\\P(6)=\frac{33}{199}\\\\\text{Dice is fair}\\\\P(1,2,3,4,5,6}=\frac{33}{199}$

→→→To check whether the result are significant or not , we will calculate standard error(e) and then z value

1.

$(e_{1})^2=(P_{1})^2+(P'_{1})^2\\\\(e_{1})^2=[\frac{28}{199}]^2+[\frac{33}{199}]^2\\\\(e_{1})^2=\frac{1873}{39601}\\\\(e_{1})^2=0.0472967\\\\e_{1}=0.217478\\\\z_{1}=\frac{P'_{1}-P_{1}}{e_{1}}\\\\z_{1}=\frac{\frac{33}{199}-\frac{28}{199}}{0.217478}\\\\z_{1}=\frac{5}{43.27}\\\\z_{1}=0.12$

→→If the value of z is between 2 and 3 , then the result will be significant at 5% level of Significance.Here value of z is very less, so the result is not significant.

2.

$(e_{2})^2=(P_{2})^2+(P'_{2})^2\\\\(e_{2})^2=[\frac{29}{199}]^2+[\frac{33}{199}]^2\\\\(e_{2})^2=\frac{1930}{39601}\\\\(e_{2})^2=0.04873\\\\e_{2}=0.2207\\\\z_{2}=\frac{P'_{2}-P_{2}}{e_{2}}\\\\z_{2}=\frac{\frac{33}{199}-\frac{29}{199}}{0.2207}\\\\z_{2}=\frac{4}{43.9193}\\\\z_{2}=0.0911$

Result is not significant.

3.

$(e_{3})^2=(P_{3})^2+(P'_{3})^2\\\\(e_{3})^2=[\frac{47}{199}]^2+[\frac{33}{199}]^2\\\\(e_{3})^2=\frac{3298}{39601}\\\\(e_{3})^2=0.08328\\\\e_{3}=0.2885\\\\z_{3}=\frac{P_{3}-P'_{3}}{e_{3}}\\\\z_{3}=\frac{\frac{47}{199}-\frac{33}{199}}{0.2885}\\\\z_{3}=\frac{14}{57.4279}\\\\z_{3}=0.24378$

Result is not significant.

4.

$(e_{4})^2=(P_{4})^2+(P'_{4})^2\\\\(e_{4})^2=[\frac{40}{199}]^2+[\frac{33}{199}]^2\\\\(e_{4})^2=\frac{3298}{39601}\\\\(e_{4})^2=0.06790\\\\e_{4}=0.2605\\\\z_{4}=\frac{P_{4}-P'_{4}}{e_{4}}\\\\z_{4}=\frac{\frac{40}{199}-\frac{33}{199}}{0.2605}\\\\z_{4}=\frac{7}{51.8555}\\\\z_{4}=0.1349$

Result is not significant.

5.

$(e_{5})^2=(P_{5})^2+(P'_{5})^2\\\\(e_{5})^2=[\frac{22}{199}]^2+[\frac{33}{199}]^2\\\\(e_{5})^2=\frac{1573}{39601}\\\\(e_{5})^2=0.03972\\\\e_{5}=0.1993\\\\z_{5}=\frac{P'_{5}-P_{5}}{e_{5}}\\\\z_{5}=\frac{\frac{33}{199}-\frac{22}{199}}{0.1993}\\\\z_{5}=\frac{11}{39.6610}\\\\z_{5}=0.2773$

Result is not significant.

6.

$(e_{6})^2=(P_{6})^2+(P'_{6})^2\\\\(e_{6})^2=[\frac{33}{199}]^2+[\frac{33}{199}]^2\\\\(e_{6})^2=\frac{2178}{39601}\\\\(e_{6})^2=0.05499\\\\e_{6}=0.2345\\\\z_{6}=\frac{P'_{6}-P_{6}}{e_{6}}\\\\z_{6}=\frac{\frac{33}{199}-\frac{33}{199}}{0.2345}\\\\z_{6}=\frac{0}{46.6655}\\\\z_{6}=0$

Result is not significant.

⇒If you will calculate the mean of all six z values, you will obtain that, z value is less than 2.So, we can say that ,outcomes are not equally likely at a 0.05 significance level.

⇒⇒Yes , as Probability of most of the numbers that is, 1,2,3,4,5,6 are different, for a loaded die , it should be equal to approximately equal to 33 for each of the numbers from 1 to 6.So, we can say with certainty that loaded die behaves differently than a fair die.

8 0

3 years ago

The distance between the cities Atlanta and Indianapolis is 540 miles. A motorcycle's speed is 48 mph. For a car it takes 2.25 h

bezimeni [28]

Answer:

Step-by-step explanation:

let they travel the distance between them alone

d = common distance traveled by both motorcycle and car = 540 miles

$t_{m}$ = time taken by motorcycle to travel distance between them

$v_{m}$ = speed of the motorcycle = 48 mph

time taken by motorcycle to travel distance between them is given as

$t_{m}$ = d/ $v_{m}$

$t_{m}$ =540/48

$t_{m}$ = 11.25 h

$t_{c}$ = time taken by car to travel distance "d" =11.25 - 2.25 = 9 h

$v_{m}$ = speed of the motorcycle

speed of motorcycle is given as

$v_{m}$ = d/ $t_{m}$

$v_{m}$ = 540/9

$v_{m}$ = 60 mph

when the car and motorcycle move towards each other, the relative speed is given as

v = $v_{m}$ + $v_{c}$

v = 60 + 48

v = 108 mph

t = time taken to meet

Time taken to meet is given as

t = d/v

t = 540/108

t = 5 h

7 0

3 years ago

Missy started with $217 in her bank account. She deposits $25.50 each week and never withdraws any money. What expression can Mi

JulijaS [17]

Answer: yep

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers