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3 years ago

AIIIIIIIIIIIII READ DIS!!!!!!!

Mathematics

1 answer:

Naddik [55]3 years ago

4 0

Answer:

Sales tax: $2.37

Subtotal: $16.95

Step-by-step explanation:

$24.95×0.095=$2.37 sales tax

$1.61÷0.095=$16.9473 or $16.95 subtotal

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The population of a small town is 23,781 and has been continuously declining at a rate of 3.2% each year what is the approximate

timurjin [86]

Same as the car one

A=Pe^(rt)
when it is a decay, make r negative
P=23781 and r=-0.032 and t=17
A=23781e^(-0.032*17)
A=23781e^(-0.554)
A=13803.01
so about 13803 people is the population after 17 years

8 0

4 years ago

Read 2 more answers

Gerardo made the following scores on his first semester science tests: 80, 85, 85, 87, 97, 100. Which measure would he use in su

Elza [17]

The mean is 89 the median is 86 the mode is 85 Hope this helps!

(technically the mean is like 88.7 so I just rounded up to get 89)

4 0

3 years ago

HELP ME !! AND EXPLAIN HOW YOU GOT IT.

Juli2301 [7.4K]

Answer: G = 50°, I = 50°, H = 80°

Step-by-step explanation:

First and foremos, the sum of three angles of a triangle is 180°

2x + 2x + (4x - 20) = 180°

8x - 20 = 180°

8x = 180° + 20 (adding 20 on both sides)

8x = 200°

x = 200/8

x = 25

The value of the angles can be found by replacing the value of x with 25

2x = 2 x 25 = 50°

Therefore, the angle of two angles are 50°

And the other is (4 x 25) -20 = 80°

4 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 $\binom{r}{r-1}$