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2 years ago

Matthew has 3 1/4 gallons of lemonade. He sells 4/8 of the lemonade. How many gallons of lemonade does Matthew have left?

Mathematics

1 answer:

Yanka [14]2 years ago

8 0

Answer:

He would have 11/4 gallons of lemonade or 2 3/4

Step-by-step explanation:

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What is the equation of the line that has a y intercept of (0,-3) and passes the point (-9,-9)

zavuch27 [327]

Answer:

y = 2/3x - 3

Step-by-step explanation:

First calculate the slope: y2 - y1 divided by x2 - x1

Slope: -6/-9 = 2/3

The y-intercept is -3

The equation of a line: y = mx + b

y = 2/3x - 3

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2 years ago

Give work for equations following: -x+2y<-4 2x+3y>6 thanks sooooo much!!!!

Schach [20]

Answer: See below

Step-by-step explanation:

A) $-x+2y < -4$

$-x+2y-2y < -4-2y$

$-x < -4-2y$

Multiply both sides by -1 and reverse the inequality:

$\left(-x\right)\left(-1\right) > -4\left(-1\right)-2y\left(-1\right)$

$x > 4+2y$

B) $2x+3y > 6$

$2x+3y-3y > 6-3y$

$2x > 6-3y$

$\frac{2x}{2} > \frac{6}{2}-\frac{3y}{2}$

$x > \frac{6-3y}{2}\\x > 3-\frac{3y}{2}$

3 0

2 years ago

During a science experiment the temperature of a solution in beaker 1 was 5 degrees below zero. The temperature of a solution in

liberstina [14]

The temperature in beaker 2 was 5°

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2 years ago

Read 2 more answers

A statistics textbook chapter contains 60 exercises, 6 of which are essay questions. A student is assigned 10 problems. (a) What

Scilla [17]

Answer:

a) P=0.3174

b) P=0.4232

c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

$P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}$

a) What is the probability that none of the questions are essay?

$P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174$

b) What is the probability that at least one is essay?

$P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232$

c) What is the probability that two or more are essay?

$P(X\geq2)=1-(P(0)+P(1))=1-(0.3174+0.4232)=1-0.7406=0.2594$

8 0

2 years ago

6m+7n+5m−3n simpified

diamong [38]

Answer:

11m + 4n

Step-by-step explanation:

when two numbers with the same variable are added together or subtracted the result is just the numbers added or subtracted. So you can add 6m and 5m to get 11m. and you can subtract 3n from 7n to get 4n

11m + 4n

6 0

2 years ago