Simplify 659/8 What is the answer rounded to the nearest whole number

Please help!! ill give brainliest!!! this is worth 30 points!!!!!!!!

Svetlanka [38]

Answer: a. 45 = k

b. d/t = 45

c. d = 45t

Step-by-step explanation:

The constant of proportionality (k) equals y divided by x if it is a direct variation. In the given problem, y is represented by distance (d) and x is represented by time (t).

$a)\quad \dfrac{d}{t}=k\\\\\\.\qquad \dfrac{90}{2}=k\\\\\\.\qquad 45=k\\\\\\b)\quad \dfrac{d}{t}=k\\\\\\.\ \quad \dfrac{d}{t}=45\\\\\\c)\quad \text{multiply both sides by t:}\\.\qquad d=45t$

5 0

3 years ago

Erica bought 3 1/2 yards of fabric. If she uses 2/3 of fabric to make a curtain, how much will she have left?

fgiga [73]

3 1/2 - (2/3)(3 1/2) =

7/2 - (2/3)(7/2) =

7/2 - 7/3 =

21/6 - 14/6 =

7/6 or 1 1/6 yds

3 0

3 years ago

4,-4 on the coordinate plane would 4 go on the x-axis or y-axis bc im du mb :)

ANTONII [103]

4 would go on the x axis because it always starts like this (x, y)

5 0

3 years ago

Round 34,689.37 to: a. the nearest tenths: b. the nearest thousands:

Irina18 [472]

Answer:

A. 34,689.370

b.35,000.00

Step-by-step explanation:

3 0

4 years ago

Willow Brook National Bank operates a drive-up teller window that allows customers to complete bank transactions without getting

slega [8]

Answer:

(a) λ = 2.

(b) P (X = 0) = 0.1353; P (X = 1) = 0.2706;

P (X = 2) = 0.2706; P (X = 3) = 0.1804

(c) P (Delay Problems) = 0.1431.

Step-by-step explanation:

Let X = number of arrivals at the drive-up teller window.

The average number of arrivals at the drive-up teller window per minute is,

p = 0.4 customers/ minute.

(1)

Compute the expected number of customers at the drive-up teller window in n = 5 minutes as follows:

$E(X)=\lambda\\=np\\=5\times 0.4\\=2$

Thus, the mean number of customers that will arrive in a five-minute period is λ = 2.

(2)

The random variable X follows a Poisson distribution with parameter λ = 2.

The probability mass function of X is:

$P(X=x)=\frac{e^{-2}2^{x}}{x!};\ x=0,1,2,3...$

Compute the probability of exactly 0 arrivals in 5 minutes as follows:

$P(X=0)=\frac{e^{-2}2^{0}}{0!}=\frac{0.1353\times 1}{1}=0.1353$

Compute the probability of exactly 1 arrivals in 5 minutes as follows:

$P(X=1)=\frac{e^{-2}2^{1}}{1!}=\frac{0.1353\times 2}{1}=0.2706$

Compute the probability of exactly 2 arrivals in 5 minutes as follows:

$P(X=2)=\frac{e^{-2}2^{2}}{2!}=\frac{0.1353\times 4}{2}=0.2706$

Compute the probability of exactly 3 arrivals in 5 minutes as follows:

$P(X=3)=\frac{e^{-2}2^{3}}{3!}=\frac{0.1353\times 8}{6}=0.1804$

Thus, the values are:

P (X = 0) = 0.1353

P (X = 1) = 0.2706

P (X = 2) = 0.2706

P (X = 3) = 0.1804

(3)

Delays occur in the service time if there are more than three customers arrive during any five-minute period.

Compute the probability that there are more than 3 customers as follows:

P (X > 3) = 1 - P (X ≤ 3)

$=1-\sum\limits^{3}_{x=0}{\frac{e^{-2}2^{x}}{x!}}\\=1-(0.1353+0.2706+0.2706+0.1804)\\=1-0.8569\\=0.1431$

Thus, the probability that delays will occur is 0.1431.

3 0

4 years ago