what is the coefficient of x^2y^3 in the expansion of (2x+y)^5? A. 2 B. 5 C. 40 D. 80 E. it does not exist

Erry has taken a random sample of students and determined the number of electives that each student in his sample took last year

Cerrena [4.2K]

The mean should be about the same, because the sample is random

6 0

3 years ago

Read 2 more answers

I need help with this problem

-BARSIC- [3]

Regular is
$97.99 after sale price is 83.30
97.99- 83.30= 14.69 is the discount
You can do it same way with B C and D

6 0

4 years ago

Jack received $300 for his 13th birthday. If he saves it in a bank account with 8% interest compounded yearly, how much money wi

kari74 [83]

300(1.08)^5

440.798423

7 0

3 years ago

A particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams. The heavies

RoseWind [281]

Answer:

The heaviest 5% of fruits weigh more than 747.81 grams.

Step-by-step explanation:

We are given that a particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams.

Let X = <u><em>weights of the fruits</em></u>

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean weight = 733 grams

$\sigma$ = standard deviation = 9 grams

Now, we have to find that heaviest 5% of fruits weigh more than how many grams, that means;

P(X > x) = 0.05 {where x is the required weight}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-733}{9}$ ) = 0.05

P(Z > $\frac{x-733}{9}$ ) = 0.05

In the z table the critical value of z that represents the top 5% of the area is given as 1.645, that means;

$\frac{x-733}{9}=1.645$

${x-733}}=1.645\times 9$

$x = 733 + 14.81$

x = 747.81 grams

Hence, the heaviest 5% of fruits weigh more than 747.81 grams.

8 0

3 years ago

A certain car depreciates such that its value at the end of each year is p % less than its value at the end of the previous year

icang [17]

Answer:

b(b/a)^2

Step-by-step explanation:

Given that the value of the car depreciates such that its value at the end of each year is p % less than its value at the end of the previous year and that car was worth a dollars on December 31, 2010 and was worth b dollars on December 31, 2011, then

b = a - (p% × a) = a(1-p%)

b/a = 1 - p%

p% = 1 - b/a = (a-b)/a

Let the worth of the car on December 31, 2012 be c

then

c = b - (b × p%) = b(1-p%)

Let the worth of the car on December 31, 2013 be d

then

d = c - (c × p%)

d = c(1-p%)

d = b(1-p%)(1-p%)

d = b(1-p%)^2

d = b(1- (a-b)/a)^2

d = b((a-a+b)/a)^2

d = b(b/a)^2 = b^3/a^2

The car's worth on December 31, 2013 = b(b/a)^2 = b^3/a^2

4 0

4 years ago