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3 years ago

8

Erin is buying produce at a store. She buys c cucumbers at $0.69 each and a apples at $0.89 each. What does the expression 0.69c

+ 0.89a represent?
The expression represents the
(select)
.

1 answer:

Basile [38]3 years ago

6 0

Answer:

It represents the total amount that Erin is going to spend at the store.

Step-by-step explanation:

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the 30 year olds are more likely to pick chocolate as their favorite ice cream because the fraction of people who like chocolate is 14/20 where the other 3 are 4/20 and 2/20. this can be converted into percentages as well by multipling by 5. 70% of people who are in their 30s like chocolate opposed to the 20% and 10%.

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Ased only on the information given in the diagram, which congruence theorems could be given as reasons why JOY ~ LIM? Check all

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3 years ago

Marissa is drawing coins from a bag that contains 5 pennies (yellow), 4 nickels (green), 5 dimes

Anna007 [38]

Answer:

$\frac{4}{16}=\frac{1}{4}=0.25=25\%$

Step-by-step explanation:

Total number of coins in the bag = 5 + 4 + 5 + 2 = 16 coins

Number of nickels in the bag = 4

We have to find the probability that Marissa draws a nickels from the bag.

Probability is defined as the ratio of Number of "Favorable Outcomes" to "Total number of possible outcomes"

In this case total number of possible outcomes is the total number of coins which is 16.

Favorable/Desires outcome is drawing a Nickle from the bag which are 4 in number. So, number of favorable outcomes is 4

Therefore, the probability that she will draw a nickel = $\frac{4}{16}=\frac{1}{4}=0.25=25\%$

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3 years ago

Customers arrive at a grocery store at an average of 2.1 per minute. Assume that the number of arrivals in a minute follows the

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: $\lambda=2.1$

$P(2)=\frac{2.1^2 \cdot e^{-2.1}}{2}\\\\P(2)=0.270$

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