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3 years ago

I really need someone to explain to me on how to solve these!! :(

Mathematics

1 answer:

HACTEHA [7]3 years ago

5 0

Answer:

3.7%

Step-by-step explanation:

(p)principal=$3400

(t)time=9months=9÷12year=3÷4year

(i)interest=$94.50

(r)rate=?

we have

r%=i/(pt)

r%=94.50÷(3400×3÷4)

r÷100=94.50÷2550

r=0.37×100

r=3.7

required rate is 3/7%

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maks197457 [2]

Minimum = 827
Q1 = 886
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3 years ago

If x= 3, what is the value of 9x + 25?<br> A. 102<br> B. 252<br> C. 66<br> D. 52

Scrat [10]

Answer:D

Step-by-step explanation:9*3=27

27+25=52

4 0

3 years ago

Read 2 more answers

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