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3 years ago

How do I solve -6= -9/u

Mathematics

1 answer:

Aleonysh [2.5K]3 years ago

5 0

U= 9 divide by 6

U=9/6

=3/2

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lisa has played in 6 soccer matches. Her brother josh has played in 18 soccer matches. lisa saay josh has played in 12 times as

Basile [38]

Answer:

Josh has played 12 matches more, not 12 times as many as her.

Step-by-step explanation:

Lisa said times, making the statement incorrect.

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3 years ago

Caleb spent 34.95 on 5 pounds of coffee. how much was the coffee per pound?

horsena [70]

It would be $6.99. All you have to do is divide the total (34.95) by pounds of coffee (5). 34.95 / 5 will give you 6.99! You can check this by doing 6.99 * 5, which would again give you 34.95. Contact me if you need any more help. :^)

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3 years ago

What would the answer be because I already tried everything and I just can’t get it

Mrac [35]

Answer:

e = 56

Step-by-step explanation:

Use cross multiplication.

28 * 14 = 7e

Simplify

392 = 7e

Divide each side by 7

e = 56.

5 0

3 years ago

Read 2 more answers

The following patterns are made using small squares

Alex787 [66]

Answer:

stuck on that too

Step-by-step explanation:

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3 years ago

An apartment complex rents an average of 2.3 new units per week. If the number of apartment rented each week Poisson distributed

masya89 [10]

Answer:

$P(X\leq 1) = 0.331$

Step-by-step explanation:

Given

Poisson Distribution;

Average rent in a week = 2.3

Required

Determine the probability of renting no more than 1 apartment

A Poisson distribution is given as;

$P(X = x) = \frac{y^xe^{-y}}{x!}$

Where y represents λ (average)

y = 2.3

<em>Probability of renting no more than 1 apartment = Probability of renting no apartment + Probability of renting 1 apartment</em>

<em />

Using probability notations;

$P(X\leq 1) = P(X=0) + P(X =1)$

Solving for P(X = 0) [substitute 0 for x and 2.3 for y]

$P(X = 0) = \frac{2.3^0 * e^{-2.3}}{0!}$

$P(X = 0) = \frac{1 * e^{-2.3}}{1}$

$P(X = 0) = e^{-2.3}$

$P(X = 0) = 0.10025884372$

Solving for P(X = 1) [substitute 1 for x and 2.3 for y]

$P(X = 1) = \frac{2.3^1 * e^{-2.3}}{1!}$

$P(X = 1) = \frac{2.3 * e^{-2.3}}{1}$

$P(X = 1) =2.3 * e^{-2.3}$

$P(X = 1) = 2.3 * 0.10025884372$

$P(X = 1) = 0.23059534055$

$P(X\leq 1) = P(X=0) + P(X =1)$

$P(X\leq 1) = 0.10025884372 + 0.23059534055$

$P(X\leq 1) = 0.33085418427$

$P(X\leq 1) = 0.331$

Hence, the required probability is 0.331

6 0

3 years ago