8 strawberries 5 bananas 3 oranges <br><br> Strawberries to bananas and oranges

The initial number of veiws for a reader board was 25. The number of views is growing at a rate of 18% per week.

Ugo [173]

Based on the given values above, we will be able to determine the expected number of viewers after four weeks by just multiplying the number of viewers to 18% per week. So the initial is 25 for this next. So for the first week, additional of 4.5 viewers. So the total would be 29.5 Second week, additional of 5.31, and the total would be 34.81. Third week, additional 6.27, the total would be 41.08. And on the fourth week, additional of 7.39, so the expected number of viewers would be 48.47. To round this off to the nearest whole number, we can get 48 viewers. Hope this helps.

3 0

2 years ago

Read 2 more answers

How many times greater is the value of 4 in 547 than the value of the 4 and 84

sesenic [268]

Answer:

10

Step-by-step explanation:

How many times greater is the value of 4 in 547 than the value of the 4 and 84?

547: 500 + 40 + 7

5 is in the hundreds place

4 is in the tens place

7 is in the ones place

84: 80 + 4

8 is in the tens place

4 is in the ones place

The value of 4 in 547 is 10 times greater than the value of 4 in 84.

Since the 4 in 547 is in the tens place, while the 4 in 84 is in the ones place.

Also, 4 * 10 = 40

Hope this helps!

6 0

2 years ago

Norma has a credit card that uses the adjusted balance method. For the first

Stella [2.4K]

Answer:

50

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Suppose that 50% of all young adults prefer McDonald's to Burger King when asked to state a preference. A group of 12 young adul

ddd [48]

Answer:

a) 0.194 = 19.4% probability that more than 7 preferred McDonald's

b) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

c) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

Step-by-step explanation:

For each young adult, there are only two possible outcomes. Either they prefer McDonalds, or they prefer burger king. The probability of an adult prefering McDonalds is independent from other adults. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

50% of all young adults prefer McDonald's to Burger King when asked to state a preference.

This means that $p = 0.5$

12 young adults were randomly selected

This means that $n = 12$

(a) What is the probability that more than 7 preferred McDonald's?

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.121$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.054$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.016$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.003$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.000$

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.121 + 0.054 + 0.016 + 0.003 + 0.000 = 0.194$

0.194 = 19.4% probability that more than 7 preferred McDonald's

(b) What is the probability that between 3 and 7 (inclusive) preferred McDonald's?

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.054$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.121$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.193$

$P(X = 6) = C_{12,6}.(0.5)^{6}.(0.5)^{6} = 0.226$

$P(X = 7) = C_{12,7}.(0.5)^{7}.(0.5)^{5} = 0.193$

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.054 + 0.121 + 0.193 + 0.226 + 0.193 = 0.787$

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

(c) What is the probability that between 3 and 7 (inclusive) preferred Burger King?

Since $p = 1-p = 0.5$ , this is the same as b) above.

So

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

7 0

2 years ago

Please help out !!!!

evablogger [386]

Answer:

x = 8

Step-by-step explanation:

you have similar triangles here

so 2 is to 4 as 4 is to x

2/4 = 4/x

x = 8

8 0

3 years ago

8 strawberries 5 bananas 3 oranges Strawberries to bananas and oranges