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2 years ago

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Nina has been saving her earnings from her lemonade stand. Nina has 70 dollars to spend. If she buys a new hat for 15 dollars, h

ow many bows can she buy with the remaining money if they cost 4 dollars each?

1 answer:

natta225 [31]2 years ago

4 0

Answer:

13 bows

Step-by-step explanation:

Given that she has 70 dollars, if she buys a new hat for 15 dollars then the amount she would have left

= 70 - 15

= 55 dollars

If she is to buy bows with the remaining amount and each cost 4 dollars then the number of bows that can be bought

= 55/4

= 13 3/4

This means that she can buy 13 bows and will have 3 dollars left as the 13 will cost 52 dollars (13 * 4).

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Suppose the number of children in a household has a binomial distribution with parameters n=12n=12 and p=50p=50%. Find the proba

nadya68 [22]

Answer:

a) 20.95% probability of a household having 2 or 5 children.

b) 7.29% probability of a household having 3 or fewer children.

c) 19.37% probability of a household having 8 or more children.

d) 19.37% probability of a household having fewer than 5 children.

e) 92.71% probability of a household having more than 3 children.

Step-by-step explanation:

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.

In this problem, we have that:

$n = 12, p = 0.5$

(a) 2 or 5 children

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

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.1934$

$p = P(X = 2) + P(X = 5) = 0.0161 + 0.1934 = 0.2095$

20.95% probability of a household having 2 or 5 children.

(b) 3 or fewer children

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

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

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0002 + 0.0029 + 0.0161 + 0.0537 = 0.0729$

7.29% probability of a household having 3 or fewer children.

(c) 8 or more children

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)$

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

$P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.1208$

$P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.0537$

$P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.0161$

$P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.0029$

$P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.0002$

$P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.1208 + 0.0537 + 0.0161 + 0.0029 + 0.0002 = 0.1937$

19.37% probability of a household having 8 or more children.

(d) fewer than 5 children

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

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

$P(X = 0) = C_{12,0}.(0.5)^{0}.(0.5)^{12} = 0.0002$

$P(X = 1) = C_{12,1}.(0.5)^{1}.(0.5)^{11} = 0.0029$

$P(X = 2) = C_{12,2}.(0.5)^{2}.(0.5)^{10} = 0.0161$

$P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.0537$

$P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.1208$

$P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0002 + 0.0029 + 0.0161 + 0.0537 + 0.1208 = 0.1937$

19.37% probability of a household having fewer than 5 children.

(e) more than 3 children

Either a household has 3 or fewer children, or it has more than 3. The sum of these probabilities is 100%.

From b)

7.29% probability of a household having 3 or fewer children.

p + 7.29 = 100

p = 92.71

92.71% probability of a household having more than 3 children.

5 0

3 years ago

Point J(-2, 1) and point K(4, 5) form the line segment jk. for the point p that partitions jk in the ratio 3:7 what is the y coo

kumpel [21]

Answer:

$\frac{11}{5}$

Step-by-step explanation:

Using the section formula

$y_{P}$ = $\frac{3(5)+7(1)}{3+7}$ = $\frac{15+7}{10}$ = $\frac{22}{10}$ = $\frac{11}{5}$

7 0

2 years ago

Talib is trying to find the inverse of the function to the right. His work appears beneath it. Is his work correct? Explain your

Rasek [7]

To find the inverse of a function, we make the independent variable the subject of the formula.

Thus, the inverse of the given function is evaluated as follows.
$f(x)=-8x+4 \\ y=-8x+4 \\ -8x=y-4 \\ x= \frac{y-4}{-8} \\ f^{-1}(x)=\frac{x-4}{-8}$

From the work show, it can be seen that Talib's work is correct.

4 0

3 years ago

Read 2 more answers

Given csc A= 37/12 and that angle A is in Quadrant I, find the exact value of sec A in

vredina [299]

Answer:

Step-by-step explanation:

Below is the pic of how this would be set up in order to determine what it is you are looking for. The angle is set in QI, and since csc A is the reciprocal of sin, the ratio is hypotenuse over side opposite. Solve for the missing side using Pythagorean's Theorem:

$37^2=12^2+b^2$ and

1369 = 144 + b² and

1225 = b² so

b = 35

The sec ratio is the reciprocal of cos, so if cos is adjacent over hypotenuse, the sec is hypotenuse over adjacent, which is 37/35

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3 years ago

Please help fast! (see picture)

valentina_108 [34]

$m = \frac{y2 - y1}{x2 - x1} \\ m(x2 - x1) = y2 - y1 \\ m(x2 - x1) + y1 = y2$

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2 years ago