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3 years ago

Can sumone do this plzz

Mathematics

1 answer:

Alex17521 [72]3 years ago

7 0

Answer:

1. $\frac{x}{15}$ = 0.85

2. x = 12.75

Step-by-step explanation:

First, we do 12 divided by 15, which is 0.80 (80%). The question is asking how much consecutive free throws is needed to raise it to 85%.

So, and equation we make is:

$\frac{x}{15}$ = 0.85, where x represents the number of free throws.

Now, we solve.

Multiply each side by 15 to solve for x.

x = 12.75

12.75 consecutive free throws is needed to raise the percent to 85%.

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Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand

dybincka [34]

Answer: Part a) $P(a)=\frac{1}{\binom{r+w}{r}}$

part b) $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}$

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = $P(E)=\frac{Favourablecases}{TotalCases}$

For part a)

Probability that a red ball is drawn in first attempt = $P(E_{1})=\frac{r}{r+w}$

Probability that a red ball is drawn in second attempt= $P(E_{2})=\frac{r-1}{r+w-1}$

Probability that a red ball is drawn in third attempt = $P(E_{3})=\frac{r-2}{r+w-1}$

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = $P(E_{i})=\frac{r-i}{r+w-i}$

Thus the probability that events $E_{1},E_{2}....E_{i}$ occur in succession is

$P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...$

Thus $P(E)$ = $\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!$

Thus our probability becomes

$P(E)=\frac{1}{\binom{r+w}{r}}$

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at $(r+1)^{th}$ draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals $\binom{r}{r-1}$

Thus the probability becomes $P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}$

Thus required probability of case b becomes $P(E)+ P(E_{i})$

= $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\$

7 0

4 years ago

Write a sentence of the form “–––––––––––––– is a function of –––––––.”

KATRIN_1 [288]

Answer:

my mom is a function of disappointment

Step-by-step explanation:

why? becuase shes mean

3 0

3 years ago

How do I simplify these expressions? 5(x+8)+3 3.6x-7-5.1x 4+8x+2.2-10x

Aleks04 [339]

Answer:

1. 5x +43

2. -1.5x - 7

3. 6.2 - 2x

Step-by-step explanation:

Equation 1:

5 (x+8) +3

First, we can distribute the 5 to the (x+8) and get 5x + 40. Distributing is when we multiply the 5 by the first number (x) and then by the second number (8) Because they aren't like terms (don't both have x's) we cannot combine then and must keep them separated by a subtraction sign

Now we have: 5x + 40 + 3

Next, we can combine the like terms. This means that any that have the same variable can be combined. So, the 5x has no other x's so he has to stay how he is. The 40 and the 3, however, can be added together to get 43.

Our finished equation is: 5x + 43

Equation 2:

3.6x - 7 - 5.1x

First, we can combine like terms as we learned in the last problem. This would be our x's since we have multiple.

We can add 3.6x and -5.1x and get -1.5x

Now we have: -1.5x - 7

Equation 3:

4 + 8x + 2.2 - 10x

We can start with either the numbers with x's or without but I'll just do the x's. So we have 8x and -10x. Adding these together would get us -2x.

Next, we can combine 4 and 2.2 and get 6.2.

Now, putting these back into our equation would look like this:

6.2 - 2x

I'm not sure how much my explanations helped, but I hope you understand!!

8 0

3 years ago

Convert 0.656 to a fraction. A. 21⁄32 B. 22⁄30 C. 20⁄33 D. 23⁄12

OlgaM077 [116]

The answer is A. 21/32

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4 years ago

Read 2 more answers

Pick three numbers to make a fact family when write each fact family

Hunter-Best [27]

Answer: Maybe it is 2,4

Step-by-step explanation:

A fact family is a group of math facts using the same numbers. In the case of addition and subtraction, you use three numbers and get four facts.

For example, you can form a fact family using the three numbers 3, 10, and 13: 10 + 3 = 13, 3 + 10 = 13, 13 − 10 = 3, and 13 − 3 = 10.

Addition facts − Here the sum of two of the three given numbers is the remaining third number. This sum is expressed in two possible orders. Thus, there are two addition facts for any three given numbers

For 3, 10 and 13, the two addition facts are:

3 + 10 = 13;

10 + 3 = 13

Subtraction facts − Here the difference of two of the three given numbers is the remaining third number. There are two such differences. Thus, there are two subtraction facts for any three given numbers.

For 3, 10 and 13, the two subtraction facts are:

13 − 10 = 3;

13 − 3 = 10

We use fact families to reinforce the connection between addition and subtraction, and to help students learn and memorize the basic addition & subtraction facts.

Problem 1:

Make four related number sentences using each set of 3 numbers.

10, 7, 3:

There should be two additions and two subtractions.

Solution

10, 7, 3:

Step 1:

The addition facts using above three numbers are:

7 + 3 = 10;

3 + 7 = 10;

Step 2:

The subtraction facts using above three numbers are:

10 – 7 = 3;

10 – 3 = 7

Problem 2:

Make four related number sentences using each set of 3 numbers.

13, 8, 5

There should be two additions and two subtractions.

Solution

13, 8, 5

Step 1:

The addition facts using above three numbers are:

5 + 8 = 13;

8 + 5 = 13;

Step 2:

The subtraction facts using above three numbers are:

13 – 5 = 8;

13 − 8 = 5

6 0

4 years ago