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vesna_86 [32]
3 years ago
15

In the first quadrant, you start at (7,4)

Mathematics
1 answer:
Xelga [282]3 years ago
6 0

Answer:

7

Step-by-step explanation:

From O(7,4), when you move a point 4 units left, abscissa decreases by 4,i.e it becomes 3 and when you move a point 3 units up, ordinate increases by 3,i.e it becomes 7.

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Harlamova29_29 [7]

Answer:

o aao is the answer

Step-by-step explanation:

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42300000

Step-by-step explanation:

Use known fact

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1\ km^2=1\ km\times 1\ km=1000\ m\times 1000\ m=1000000\ m^2

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42.3\ km^2=42.3\times 1000000\ m^2=42300000\ m^2

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Due to a manufacturing error, 2 cans of regular soda were accidentally filled with diet soda and placed into a24 -pack.
Gennadij [26K]

Answer:

a) 1/48 or 2.0833%

b) 46/48 or 95.8333%

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5 0
3 years ago
F(-5) if(x) - 4x + 2
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In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist
frosja888 [35]

Answer:

a) 0.9920 = 99.20% probability that 15 or more will live beyond their 90th birthday

b) 0.2946  = 29.46% probability that 30 or more will live beyond their 90th birthday

c) 0.6273 = 62.73% probability that between 25 and 35 will live beyond their 90th birthday

d) 0.0034 = 0.34% probability that more than 40 will live beyond their 90th birthday

Step-by-step explanation:

We solve this question using the normal approximation to the binomial distribution.

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 distributions 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)}.

In this problem, we have that:

Sample of 723, 3.7% will live past their 90th birthday.

This means that n = 723, p = 0.037.

So for the approximation, we will have:

\mu = E(X) = np = 723*0.037 = 26.751

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{723*0.037*0.963} = 5.08

(a) 15 or more will live beyond their 90th birthday

This is, using continuity correction, P(X \geq 15 - 0.5) = P(X \geq 14.5), which is 1 subtracted by the pvalue of Z when X = 14.5. So

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

Z = \frac{14.5 - 26.751}{5.08}

Z = -2.41

Z = -2.41 has a pvalue of 0.0080

1 - 0.0080 = 0.9920

0.9920 = 99.20% probability that 15 or more will live beyond their 90th birthday

(b) 30 or more will live beyond their 90th birthday

This is, using continuity correction, P(X \geq 30 - 0.5) = P(X \geq 29.5), which is 1 subtracted by the pvalue of Z when X = 29.5. So

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

Z = \frac{29.5 - 26.751}{5.08}

Z = 0.54

Z = 0.54 has a pvalue of 0.7054

1 - 0.7054 = 0.2946

0.2946  = 29.46% probability that 30 or more will live beyond their 90th birthday

(c) between 25 and 35 will live beyond their 90th birthday

This is, using continuity correction, P(25 - 0.5 \leq X \leq 35 + 0.5) = P(X 24.5 \leq X \leq 35.5), which is the pvalue of Z when X = 35.5 subtracted by the pvalue of Z when X = 24.5. So

X = 35.5

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

Z = \frac{35.5 - 26.751}{5.08}

Z = 1.72

Z = 1.72 has a pvalue of 0.9573

X = 24.5

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

Z = \frac{24.5 - 26.751}{5.08}

Z = -0.44

Z = -0.44 has a pvalue of 0.3300

0.9573 - 0.3300 = 0.6273

0.6273 = 62.73% probability that between 25 and 35 will live beyond their 90th birthday.

(d) more than 40 will live beyond their 90th birthday

This is, using continuity correction, P(X > 40+0.5) = P(X > 40.5), which is 1 subtracted by the pvalue of Z when X = 40.5. So

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

Z = \frac{40.5 - 26.751}{5.08}

Z = 2.71

Z = 2.71 has a pvalue of 0.9966

1 - 0.9966 = 0.0034

0.0034 = 0.34% probability that more than 40 will live beyond their 90th birthday

6 0
3 years ago
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