Khan academy use Pythagorean theorem to find isosceles triangle size length

Swimming Pool On a certain hot summer's day, 345 people used the public swimming pool. The daily prices are $1.50 for children a

Natasha2012 [34]

Answer:

143

Step-by-step explanation:

just took the test

6 0

3 years ago

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How do you find the percent of an equation??

vlabodo [156]

<span>15 is 10% what number? in equation form is 15/10% = X.We need to first convert the percentage to an equivalent decimal to do our calculation.<span>Converting a percentage to a decimal we remove the percentage sign and divide by 100.</span></span>

3 0

3 years ago

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Two life insurance policies, each with a death benefit of 10,000 and a one-time premium of 500, are sold to a couple, one for ea

Kazeer [188]

Answer:896.9

Step-by-step explanation:

Let x denotes excess premium over claims

$E\left ( x|husband survives\right )$ , There are two possibilities

(i)Only husband survives

This can be possible with a possibility of 0.01

Claims=10,000

Premium collected $=2\times 500=1000$

Thus x=1000-10,000=-9000

(ii)Both husband and wife survives

This can occur with a probability of 0.96

Here claims will be 0 as both survives

Premium taken=1000

thus x=1000

The probability that the husband survives is the sum of above cases

=0.96+0.01=0.97

Hence the desired conditional Expectation $E\left ( x|Husband survives\right ) =0.01\times \left ( -9000\right )+0.96\times \left ( 1000\right )=896.9$

7 0

3 years ago

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

3 years ago

A shopper buys cat food in bags of 3 lbs. Her cat eats 3/4 alb each week. How many weeks does one bag last?

iren2701 [21]

Divide 3 by 3/4 and you will get your answer. you can also multiply 3×4/3. it's the same thing

6 0

3 years ago