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2 years ago

Please lmk asapppppppp

Mathematics

1 answer:

lana66690 [7]2 years ago

5 0

Answer:

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Step-by-step explanation:

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Wendy has $27 to buy seed for her birdfeeders. Each bag of seed costs $5.

Alekssandra [29.7K]

Answer:

Step-by-step explanation:

If Wendy has $27 to buy seeds, and each seed bag costs $5, then:

5 · 5 = 25

5 · 6 = 30

27<30

27>25

So, Wendy is able to buy 5 bags and remain with 2 dollars.

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2 years ago

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A basketball player made 80 out of 100 attempted free throws. What percent of free throws was made?

Mrac [35]

Percent of free throws = (number of free throws made / total attempts) x 100

Percent = (80/100) x 100 = 80%

The answer is 80%

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2 years ago

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What is the mystery number? The greatest common factor of the mystery pair is 7. The least common multiple of the mystery pair i

xxMikexx [17]

The ANSWER to the mystery question is 50

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For a given event, what is the result of dividing the number of successful outcomes by the number of possible outcomes?

Zielflug [23.3K]

Dividing the number of successful outcomes by the number of possible outcomes is the definition of the probability of the given event.

P(E)=n(E)/n(S)

example. let the Experiment be throwing a dice. The sample space, all the possible outcomes, is the set S={1, 2, 3, 4, 5, 6 } and n(S)=6

Let the even E be: getting an even number. The successful outcomes set is E {2, 4, 6 } and n(E)=3

So probability of getting an even number is P(E)=n(E)/n(S)=3/6=0.5

Answer: Probability of the given Event

5 0

3 years ago

According to the 2011 National Survey of Fishing, Hunting, and Wildlife-Associated Recreation, there were over 71 million wildif

suter [353]

Answer:

The probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011 is P=0.412.

Step-by-step explanation:

We know the population proportion π=0.8, according to the 2011 National Survey of Fishing, Hunting, and Wildlife-Associated Recreation.

If we take a sample from this population, and assuming the proportion is correct, it is expected that the sample's proportion to be equal to the population's proportion.

The standard deviation of the sample is equal to:

$\sigma_s=\sqrt{\frac{\pi(1-\pi)}{N}}=\sqrt{\frac{0.8*0.2}{500}}=0.018$

With the mean and the standard deviaion of the sample, we can calculate the z-value for 0.79 and 0.81:

$z=\frac{p-\pi}{\sigma}=\frac{0.79-0.80}{0.018} =\frac{-0.01}{0.018} = -0.55\\\\z=\frac{p-\pi}{\sigma}=\frac{0.81-0.80}{0.018} =\frac{0.01}{0.018} = 0.55$

Then, the probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011 is:

$P(0.79$

3 0

2 years ago