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Neporo4naja [7]

3 years ago

14

Adrian earns $3600 per month. he spends 1/4 of his income on rent, 2/5 on food, 1/9 on activities, and he saves the rest. how mu

ch does he save each month?

2 answers:

arsen [322]3 years ago

7 0

To add the fractions we must make the denominators = 180.

45/180 + 72/180 + 20/180 = 137/180

Expense Total = (137/180) * 3,600 = $2,740

Savings Total = (180 - 137) / 180

Savings = (43/180) * 3,600

Savings = $860

***********DOUBLE CHECK**********************

1/4=.25 2/5 = .40 1/9 = .111111

.25 + .40 + .111111 = .761111

Expenses = .761111 * 3,600 = 2,740.00

Savings = 3,600 - 2,740.00 = 860.00

murzikaleks [220]3 years ago

4 0

he saves $860 per month

calculate each of his outgoings

rent = $\frac{1}{4}$ × $3600 = $\frac{3600}{4}$ = $900

food = $\frac{2}{5}$ × $3600 = $\frac{2(3600)}{5}$ = $1440

activities = $\frac{1}{9}$ × $3600 = $\frac{3600}{9}$ = $400

savings = $3600 - ( 900 + 1440 + 400 ) = $860

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Answer:

a = 28.14 (Estimated)

Step-by-step explanation:

7a + 50 = 247

=> 7a = 197

=> a = 28.1428571429 (Using Calculator)

=> a = 28.14 (Estimated)

Hoped this helped.

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3 years ago

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hello,

Step-by-step explanation:

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a=69(°)

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3 years ago

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Answer:

$-1\frac{1}{2}$ , $-\frac{3}{4}$ , $-\frac{37}{50}$ , $\frac{5}{4}$ , $1\frac{1}{4}$ , $2\frac{3}{4}$

Step-by-step explanation:

$1\frac{1}{4}$ , $-1\frac{1}{2}$ , $-\frac{37}{50}$ , $-\frac{3}{4}$ , $2\frac{3}{4}$ , $\frac{5}{4}$

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$-1\frac{1}{2}$ , $-\frac{37}{50}$ , $-\frac{3}{4}$

<u><em>Positive numbers:</em></u>

$1\frac{1}{4}$ , $2\frac{3}{4}$ , $\frac{5}{4}$

<em><u>Now order it on a number line:</u></em>

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3 years ago

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Black_prince [1.1K]

Answer:

a) P(2)=0.270

b) P(X>3)=0.605

c) P=0.410

Step-by-step explanation:

We know that customers arrive at a grocery store at an average of 2.1 per minute. We use the Poisson distribution:

$\boxed{P(k)=\frac{\lambda^k \cdot e^{-\lambda}}{k!}}$

a) In this case: $\lambda=2.1$

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Therefore, the probability is P(2)=0.270.

b) In this case: $\lambda=2\cdot 2.1=4.2$

$P(X>3)=1-P(X\leq 3)\\\\P(X>3)=1-\sum_{x=0}^3 \frac{4.2^x \cdot e^{-4.2}}{x!}\\\\P(X>3)=1-0.395\\\\P(X>3)=0.605$

Therefore, the probability is P(X>3)=0.605.

c) We know that two customers came in in the first minute. That is why we calculate the probability of at least 5 customers entering the other 2 minutes.

In this case: $\lambda=2\cdot 2.1=4.2$

$P(X\geq 5)=1-P(X$

Therefore, the probability is P=0.410.

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3 years ago