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Vinil7 [7]
2 years ago
14

Beth is making a rope ladder.each step of the ladder is 2 1/3 feet wide.Beth has a rope that is 21 feet long.how many steps can

she make from the rope?
Mathematics
2 answers:
butalik [34]2 years ago
8 0

Answer: She can make 9 steps from the rope.

Step-by-step explanation:

Given : The width of each step of the ladder = 2\dfrac{1}{3}\text{ feet} = \dfrac{2(3)+1}{3}=\dfrac{7}{3}\text{ feet}

The length of the rope = 21 feet

Then , the number of required steps = (  length of the rope ) ÷ (width of each step of the ladder)

=21\div\dfrac{7}{3}

=21\times\dfrac{3}{7}=3\times3=9

Hence, the number of steps can she make from the rope = 9.

Marianna [84]2 years ago
7 0

Answer:

Beth can make 9 2 1/3 foot steps

Step-by-step explanation:

21/2.3=9.1

round to 9

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Answer:

(x+1)(2x+5)

Step-by-step explanation:

f(x) = 2x² + 7x + 5

Factor the expression by grouping. First, the expression needs to be rewritten as 2x²+ax+bx+5. To find a and b, set up a system to be solved.

a+b=7

ab=2×5=10

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 10.

1,10

2,5

Calculate the sum for each pair.

1+10=11

2+5=7

The solution is the pair that gives sum 7.

a=2

b=5

2x²+7x+5 as (2x²+2x)+(5x+5).

(2x²+2x)+(5x+5)

Factor out 2x in the first and 5 in the second group.

2x(x+1)+5(x+1)

Factor out common term x+1 by using distributive property.

(x+1)(2x+5)

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A 95% confidence interval was computed using a sample of 16 lithium batteries, which had a sample mean life of 645 hours. The co
polet [3.4K]

Answer:

Step-by-step explanation:

Hello!

The mean life of 16 lithium batteries was estimated with a 95% CI:

(628.5, 661.5) hours

Assuming that the variable "X: Duration time (life) of a lithium battery(hours)" has a normal distribution and the statistic used to estimate the population mean was s Student's t, the formula for the interval is:

[X[bar]±t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }]

The amplitude of the interval is calculated as:

a= Upper bond - Lower bond

a= [X[bar]+t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }] -[X[bar]-t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }]

and the semiamplitude (d) is half the amplitude

d=(Upper bond - Lower bond)/2

d=([X[bar]+t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }] -[X[bar]-t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }] )/2

d= t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }

The sample mean marks where the center of the calculated interval will be. The terms of the formula that affect the width or amplitude of the interval is the value of the statistic, the sample standard deviation and the sample size.

Using the semiamplitude of the interval I'll analyze each one of the posibilities to see wich one will result in an increase of its amplitude.

Original interval:

Amplitude: a= 661.5 - 628.5= 33

semiamplitude d=a/2= 33/2= 16.5

1) Having a sample with a larger standard deviation.

The standard deviation has a direct relationship with the semiamplitude of the interval, if you increase the standard deviation, it will increase the semiamplitude of the CI

↑d= t_{n-1;1-\alpha /2} * ↑S/√n

2) Using a 99% confidence level instead of 95%.

d= t_{n_1;1-\alpha /2} * S/√n

Increasing the confidence level increases the value of t you will use for the interval and therefore increases the semiamplitude:

95% ⇒ t_{15;0.975}= 2.131

99% ⇒ t_{15;0.995}= 2.947

The confidence level and the semiamplitude have a direct relationship:

↑d= ↑t_{n_1;1-\alpha /2} * S/√n

3) Removing an outlier from the data.

Removing one outlier has two different effects:

1) the sample size is reduced in one (from 16 batteries to 15 batteries)

2) especially if the outlier is far away from the rest of the sample, the standard deviation will decrease when you take it out.

In this particular case, the modification of the standard deviation will have a higher impact in the semiamplitude of the interval than the modification of the sample size (just one unit change is negligible)

↓d= t_{n_1;1-\alpha /2} * ↓S/√n

Since the standard deviation and the semiamplitude have a direct relationship, decreasing S will cause d to decrease.

4) Using a 90% confidence level instead of 95%.

↓d= ↓t_{n_1;1-\alpha /2} * S/√n

Using a lower confidence level will decrease the value of t used to calculate the interval and thus decrease the semiamplitude.

5) Testing 10 batteries instead of 16. and 6) Testing 24 batteries instead of 16.

The sample size has an indirect relationship with the semiamplitude if the interval, meaning that if you increase n, the semiamplitude will decrease but if you decrease n then the semiamplitude will increase:

From 16 batteries to 10 batteries: ↑d= t_{n_1;1-\alpha /2} * S/√↓n

From 16 batteries to 24 batteries: ↓d= t_{n_1;1-\alpha /2} * S/√↑n

I hope this helps!

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