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CaHeK987 [17]
3 years ago
15

When rounded to the nearest tenth, a puppy has the weight of 8.1 pounds. which of these could be the actual weight in pounds?

Mathematics
2 answers:
8_murik_8 [283]3 years ago
7 0

Step-by-step explanation:

8.1 rounded of to the nearest 10 will be 8

TiliK225 [7]3 years ago
3 0
Do you have the answer choices ?
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What is the equation of the line with slope of ½ , and which passes through the point
Law Incorporation [45]

Answer:

y = 1/2x - 7

Step-by-step explanation:

slope intercept form:

y = mx + b

y = -5

m(slope) = 1/2

x = 4

y = mx + b

<em>substitute </em>

-5 = 1/2*4 + b

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-5 - 2 = b

-7 = b

y = 1/2x - 7

6 0
3 years ago
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aleksley [76]

Answer:

f= -14

Step-by-step explanation:

4 0
3 years ago
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-1/6 + ( -3/7) = <br><br> A. -25/42<br> B. -4/13<br> C. 11/42<br> D. 2/13
alekssr [168]

Answer: the anwser is A

Step-by-step explanation:

I used a calculator lol

6 0
3 years ago
Read 2 more answers
Because not all airline passengers show up for their reserved seat, an airline sells 125 tickets for a flight that holds only 12
melamori03 [73]

Answer:

98.75% probability that every passenger who shows up can take the flight

Step-by-step explanation:

For each passenger who show up, there are only two possible outcomes. Either they can take the flight, or they do not. The probability of a passenger taking the flight is independent from other passenger. So the binomial probability distribution is used to solve this question.

However, we are working with a large sample. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

The probability that a passenger does not show up is 0.10:

This means that the probability of showing up is 1-0.1 = 0.9. So p = 0.9

Because not all airline passengers show up for their reserved seat, an airline sells 125 tickets

This means that n = 125

Using the approximation:

\mu = E(X) = np = 125*0.9 = 112.5

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{125*0.9*0.1} = 3.354

(a) What is the probability that every passenger who shows up can take the flight

This is P(X \leq 120), so this is the pvalue of Z when X = 120.

Z = \frac{X - \mu}{\sigma}

Z = \frac{120 - 112.5}{3.354}

Z = 2.24

Z = 2.24 has a pvalue of 0.9875

98.75% probability that every passenger who shows up can take the flight

7 0
3 years ago
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