MATHEMATICS COLLEGE

A pancake recipe calls for 1 cup of flour. Derek has already used 1/4 cup of flour. How many more cups of flour does Derek need to add?

Answers

Answer 1

Answer:

Answer:

Derek needs to add 3/4 cups of flour.

Step-by-step explanation:

If Derek has already used 1/4 cup of flour, he needs 3 more cups of flour to get to 1 full cup of flour.

Answer 2

Answer:

Step-by-step explanation:A batch of cookies requires 2 1/4 cup of flour and 1 egg. You have in your kitchen 8 cups of flour and a half dozen eggs. a) How many batches of cookies

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The operator of a pumping station has observed that demand for water during early afternoon hours has an approximately exponential distribution with mean 100 cfs (cubic feet per second). (a) Find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day. (Round your answer to four decimal places.)

Answers

Answer:

The probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day is $P(Y>190)=\frac{1}{e^{(19)/(10)}}\approx 0.1496$

Step-by-step explanation:

Let Y be the water demand in the early afternoon.

If the random variable Y has density function f (y) and a < b, then the probability that Y falls in the interval [a, b] is

$P(a\leq Y \leq b)=\int\limits^a_b {f(y)} \, dy$

A random variable Y is said to have an exponential distribution with parameter $\beta > 0$ if and only if the density function of Y is

$f(y)=\left \{ {{(1)/(\beta)e^{-(y)/(\beta) }, \quad{0\:\leq \:y \:\leq \:\infty} } \atop {0}, \quad elsewhere} \right.$

If Y is an exponential random variable with parameter β, then

mean = β

To find the probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day, you must:

We are given the mean = β = 100 cubic feet per second

$P(Y>190)=\int\limits^(\infty)_(190) {(1)/(100)e^(-y/100) } \, dy$

Compute the indefinite integral $\int (1)/(100)e^{-(y)/(100)}dy$

$(1)/(100)\cdot \int \:e^{-(y)/(100)}dy\n\n\mathrm{Apply\:u \:substitution}\:u=-(y)/(100)\n\n(1)/(100)\cdot \int \:-100e^udu\n\n(1)/(100)\left(-100\cdot \int \:e^udu\right)\n\n(1)/(100)\left(-100e^u\right)\n\n\mathrm{Substitute\:back}\:u=-(y)/(100)\n\n(1)/(100)\left(-100e^{-(y)/(100)}\right)\n\n-e^{-(y)/(100)}$

Compute the boundaries

$\int _(190)^(\infty \:)(1)/(100)e^{-(y)/(100)}dy=0-\left(-\frac{1}{e^{(19)/(10)}}\right)$

$\int _(190)^(\infty \:)(1)/(100)e^{-(y)/(100)}dy=\frac{1}{e^{(19)/(10)}}\approx 0.1496$

The probability that the demand will exceed 190 cfs during the early afternoon on a randomly selected day is $P(Y>190)=\frac{1}{e^{(19)/(10)}}\approx 0.1496$

Write the ratio in fractional notation in lowest terms. 28 inches to 42 inches

Answers

28 to 42 means 28/42

We now reduce 28/42 to lowest terms.

28 ÷ 7 = 4

42 ÷ 7 = 6

We now have 4/6.

We now reduce 4/6.

4 ÷ 2 = 2

6 ÷ 2 = 3

Final answer: 2/3

P(q ÷ 3 - p ); use p = -6, and q= -3

Answers

Answer:

Step-by-step explanation:

p(q ÷ 3 - p )

p = -6

q= -3

-6(-3/3 - (-6))

-6(-1 + 6)

-6(+5)

-30

By how much does the dependent variable change in response to a change of 1 unit of the independent variable?Independent 2 4 6 8
Dependent 15 25 35 45
Select one or more:
A. 1
B. 2
C. 5
D. 10