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Gwar [14]
3 years ago
14

Find the first five terms of the sequence defined below:

Mathematics
1 answer:
Nataly_w [17]3 years ago
7 0

Answer:

d. 7, -14, 28, -56, 112

Step-by-step explanation:

f(2) = -2(7) = -14

f(3) = -2(-14) = 28

f(4) = -2(28) = -56

f(5) = -2(-56) = 112

You might be interested in
The boys had 5 4 6 gallons of water to share throughout the day. James drank 1 1 3 5 6 1 2 gallons and Simon drank 2 3 gallon. H
dsp73

Answer:

1\frac{1}{3}

Step-by-step explanation:

Total Volume of Water = 5\frac{4}{6}

Volume drunk by James =1\frac{1}{3}

Volume drunk by Simon =\frac{2}{3}

Volume drunk by Matthew = 1\frac{1}{2}

Volume drunk by Gilbert =\frac{5}{6}

Total Volume Drunk (i.e. The Sum) = 1\frac{1}{3} + \frac{2}{3} + 1\frac{1}{2} +\frac{5}{6}=\frac{13}{3}

Volume of water Left = Total Volume of Water-Total Volume Drunk

= 5\frac{4}{6}-\frac{13}{3} =\frac{4}{3}=1\frac{1}{3}

3 0
4 years ago
Read 2 more answers
Help me on this please !
inna [77]

Answer:

24.08

Step-by-step explanation:

We use the Pythagorean theorem and we get (16)^2 + (18)^2 = (x)^2

256 + 324 = x^2

x^2 = 580

x is about 24.08.

7 0
3 years ago
Solve 2y-3 = 4y + 6.<br> y=?
valentinak56 [21]

y= 3/8 OR 0.375

Step-by-step explanation:

yes

5 0
3 years ago
Read 2 more answers
1. Write out the law of cosines and find the hypotenuse of triangle with 2 sides lengths equal to three and the included angle m
LUCKY_DIMON [66]

Answer:

1.) \sqrt{13.341}  ≈ 3.652

2.) I would say something about how the A in front of cos in the equation would change to 90, rather than stay 75 (in the equation for the step by step), but it would be easier to just use the Pythagorean theorem.

Step-by-step explanation:

I think we may have the same class so hopefully this helps:

1.) a^2 = b^2 + c^2 - 2bccosA --> law of cosines formula.

   a^2 = 3^2 + 3^2 - 2(3)(3)cos(75) --> plugged in numbers; when you draw the triangle, the included angle would be A, and the opposite side would be a. B and b, and C and c are opposite each other. In this case, a is the hypotenuse.

   a^2 = 18 - 18cos75 --> in between steps.

   a^2 = 13.341 --> more simplifying.

   a(hypotenuse) = \sqrt{13.341} or 3.652 --> answer

2.) This one is just an explanation: The 75 in the equation is the given angle, which is a. If this changes, it would just change in the equation too. And obviously, if it's 90 degrees, you can just use Pythagorean theorem a^2+b^2=c^2.

Good luck! :)

   

6 0
3 years ago
Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean
Mamont248 [21]

Answer:

a) The percentage of snails that take more than 60 hours to finish is 4.75%.

b) The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c) The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d) 0% probability that a randomly-chosen snail will take more than 76 hours to finish

e) To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f) The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

\mu = 50, \sigma = 6

a. The percentage of snails that take more than 60 hours to finish is

This is 1 subtracted by the pvalue of Z when X = 60.

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

Z = \frac{60 - 50}{6}

Z = 1.67

Z = 1.67 has a pvalue 0.9525

1 - 0.9525 = 0.0475

The percentage of snails that take more than 60 hours to finish is 4.75%.

b. The relative frequency of snails that take less than 60 hours to finish is

This is the pvalue of Z when X = 60.

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

Z = \frac{60 - 50}{6}

Z = 1.67

Z = 1.67 has a pvalue 0.9525

The relative frequency of snails that take less than 60 hours to finish is 95.25%.

c. The proportion of snails that take between 60 and 67 hours to finish is

This is the pvalue of Z when X = 67 subtracted by the pvalue of Z when X = 60.

X = 67

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

Z = \frac{67 - 50}{6}

Z = 2.83

Z = 2.83 has a pvalue 0.9977

X = 60

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

Z = \frac{60 - 50}{6}

Z = 1.67

Z = 1.67 has a pvalue 0.9525

0.9977 - 0.9525 = 0.0452

The proportion of snails that take between 60 and 67 hours to finish is 4.52%.

d. The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places)

This is 1 subtracted by the pvalue of Z when X = 76.

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

Z = \frac{76 - 50}{6}

Z = 4.33

Z = 4.33 has a pvalue of 1

1 - 1 = 0

0% probability that a randomly-chosen snail will take more than 76 hours to finish

e. To be among the 10% fastest snails, a snail must finish in at most hours.

At most the 10th percentile, which is the value of X when Z has a pvalue of 0.1. So it is X when Z = -1.28.

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

-1.28 = \frac{X - 50}{6}

X - 50 = -1.28*6

X = 42.32

To be among the 10% fastest snails, a snail must finish in at most 42.32 hours.

f. The most typical 80% of snails take between and hours to finish.

From the 50 - 80/2 = 10th percentile to the 50 + 80/2 = 90th percentile.

10th percentile

value of X when Z has a pvalue of 0.1. So X when Z = -1.28.

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

-1.28 = \frac{X - 50}{6}

X - 50 = -1.28*6

X = 42.32

90th percentile.

value of X when Z has a pvalue of 0.9. So X when Z = 1.28

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

1.28 = \frac{X - 50}{6}

X - 50 = 1.28*6

X = 57.68

The most typical 80% of snails take between 42.32 and 57.68 hours to finish.

5 0
4 years ago
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