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2 years ago

Jerome deposits $3,700 in a certificate of deposit that paya on Interest, compounded

Mathematics

1 answer:

Pavel [41]2 years ago

5 0

Jerome deposits $3,700 in a certificate of deposit that paya on Interest, compounded

annually. How much interest does Jerome earn in 1 year? (

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h(n)=-13nh(n)=−13nh, left parenthesis, n, right parenthesis, equals, minus, 13, n Complete the recursive formula of h(n)h(n)h, l

mr Goodwill [35]

Answer:

h(1)=-13

$h(n)=h(n-1)-13, n\geq 2$

Step-by-step explanation:

We are given that

$h(n)=-13n$

We have to find the recursive formula of h(n).

Substitute n=1

$h(1)=-13$

n=2

$h(2)=-13-13=-26=h(1)-13$

n=3

$h(3)=-13(3)=-39=-26-13=h(2)-13$

$h(4)=-13(4)=-52=-39-13=h(3)-13$

$h(n)=h(n-1)-13$

Therefore, the recursive formula is given by

h(1)=-13

$h(n)=h(n-1)-13, n\geq 2$

5 0

3 years ago

The length of human pregnancies from conception to birth follows a distribution with mean 266 days and standard deviation 15 day

Katena32 [7]

Answer:

a) 81.5%

b) 95%

c) 75%

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 266 days

Standard Deviation, σ = 15 days

We are given that the distribution of length of human pregnancies is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(between 236 and 281 days)

$P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{281-266}{15})\\\\= P(-2 \leq z \leq 1)\\\\= P(z \leq 1) - P(z < -2)\\= 0.838 - 0.023 = 0.815 = 81.5\%$

b) a) P(last between 236 and 296)

$P(236 \leq x \leq 281)\\\\= P(\displaystyle\frac{236 - 266}{15} \leq z \leq \displaystyle\frac{296-266}{15})\\\\= P(-2 \leq z \leq 2)\\\\= P(z \leq 2) - P(z < -2)\\= 0.973 - 0.023 = 0.95 = 95\%$

c) If the data is not normally distributed.

Then, according to Chebyshev's theorem, at least $1-\dfrac{1}{k^2}$ data lies within k standard deviation of mean.

For k = 2

$1-\dfrac{1}{(2)^2} = 75\%$

Atleast 75% of data lies within two standard deviation for a non normal data.

Thus, atleast 75% of pregnancies last between 236 and 296 days approximately.

7 0

3 years ago

Help me solve this problem and explain how you got the answer plz

Vitek1552 [10]

I hope this helps you

6 0

3 years ago

The diameters of ball bearings are distributed normally. The mean diameter is 87 millimeters and the standard deviation is 6 mil

devlian [24]

Answer:

69.14% probability that the diameter of a selected bearing is greater than 84 millimeters

Step-by-step explanation:

According to the Question,

Given That, The diameters of ball bearings are distributed normally. The mean diameter is 87 millimeters and the standard deviation is 6 millimeters. Find the probability that the diameter of a selected bearing is greater than 84 millimeters.

In a set with mean and standard deviation, the Z score of a measure X is given by Z = (X-μ)/σ

we have μ=87 , σ=6 & X=84

Find the probability that the diameter of a selected bearing is greater than 84 millimeters

This is 1 subtracted by the p-value of Z when X = 84.

So, Z = (84-87)/6

Z = -3/6

Z = -0.5 has a p-value of 0.30854.

⇒1 - 0.30854 = 0.69146

0.69146 = 69.14% probability that the diameter of a selected bearing is greater than 84 millimeters.

Note- (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 p-value, we get the probability that the value of the measure is greater than X)

7 0

3 years ago

Solve 80/100 = 68/n using cross multiplication.

Karo-lina-s [1.5K]

80n = 68*100

80n = 6800

n = 6800/80

n = 85

6 0

3 years ago

Read 2 more answers