All categories

2 years ago

Given the following formula, solve for I. P = 2(1 + 6) PLEASE HELP!!!

Mathematics

1 answer:

Pepsi [2]2 years ago

5 0

Step-by-step explanation

Here ...I'm not too sure about the question as you have written

P = 2 ( 1 + 6 ) and ask to solve for L ...

.So I guess the question you have written is wrong .. ...and I guess there should be L in the place of 1.

...So let me correct the question....

P = 2 ( l + 6 )

P = 2l + 12

P - 2l = 12

- 2l = 12 - P

- 2l = - P + 12

Change the sign of both side of the equation...

2l = P + 12

divide both side by the same number...

L = ( P + 12 ) ÷ 2

$l = \frac{1}{2} p - 12 \times \frac{1}{2} \\$

Calculate the product of rational numbers...

$l = \frac{1}{2} p - 6 \\$

You might be interested in

Hopefully u can see it now sigh.

telo118 [61]

Answer:

1.39m (3 s.f.)

Step-by-step explanation:

Please see attached picture for full solution.

sinθ= opp/hyp

tanθ= opp/adj

cosθ= adj/hyp

6 0

3 years ago

Find the Domain and Range

Ksju [112]

Answer:

See explanation below

Step-by-step explanation:

Domain = $-7\leq x\leq infinity$

Range = $-1$

Domain refers to the set of x values of the graph, while range refers to the set of y values of the graph.

5 0

3 years ago

Consider the following linear equations. y = 2x + 5 y = 2x - 3 Which statement below if True?

djverab [1.8K]

Answer:

Step-by-step explanation:

One of the lines is going to have the y-intercept of -3

5 0

3 years ago

Look at the picture thats where the questions are.

Ira Lisetskai [31]

$11. \: m\angle1 = m\angle2 = 135°\\ m\angle3 = 45°\\ 12.m\angle1 = 90 °\\ m\angle2 = 25°$

3 0

3 years ago

Suppose that a random sample of size 10 from the N(µ, 9) distribution resulted in a sample mean of x = 2.7. Give an 85% condiden

stellarik [79]

Answer:

a) $CI: 1.3$

b) $CI: 2.5$

c) $CI: 2.0$

Step-by-step explanation:

a) We have a sample of size n=10 and a sample mean of x=2.7.

The σ² of the population is known and is σ²=9 or σ=3.

For a CI of 85%, the z-value is 1.44.

Then the lower and upper limits of the CI are:

$LL=M-z*\frac{\sigma}{\sqrt{n} } =2.7-1.44*\frac{3}{\sqrt{10} }=2.7-1.44*\frac{3}{3.162} =2.7-1.4=1.3\\\\UL=M+z*\frac{\sigma}{\sqrt{n} } =2.7+1.4=4.1$

The 85% confidence interval for the mean is defined by the values LL=1.3 and UL=4.1.

$CI: 1.3$

b) In this case, the variance of the population is unknown.

We have to estimate the variance of the population from the variance of the sample.

Sample data:

n = 100

x(mean) = 2.7

s² = 1.1

The CI is defined as

$x-t*\frac{s}{\sqrt{n} }$

The value of t depends of the degrees of freedom and the percentage of confidence.

In this case, the degrees of freedom are n-1=100-1=99 and the CI is of 90%.

We look up in a t-table and the t-value for this conditions is 1.6604.

We can now calculate the CI

$LL: x-t*\frac{s}{\sqrt{n} }=2.7-1.6604*\frac{\sqrt{1.1} }{\sqrt{100} } =2.7-0.2=2.5\\\\UL: x+t*\frac{s}{\sqrt{n} }=2.7+0.2=2.9$

The 90% confidence interval for the mean is defined by the values LL=2.5 and UL=2.9.

$CI: 2.5$

c) In this case, the variance of the population is unknown.

We have to estimate the variance of the population from the variance of the sample.

Sample data:

n = 10

x(mean) = 2.7

s² = 1.1

The CI is defined as

$x-t*\frac{s}{\sqrt{n} }$

The value of t depends of the degrees of freedom and the percentage of confidence.

In this case, the degrees of freedom are n-1=10-1=9 and the CI is of 95%.

We look up in a t-table and the t-value for this conditions is 2.2622.

We can now calculate the CI

$LL: x-t*\frac{s}{\sqrt{n} }=2.7-2.2622*\frac{\sqrt{1.1} }{\sqrt{10} } =2.7-0.7=2.0\\\\UL: x+t*\frac{s}{\sqrt{n} }=2.7+0.7=3.4$

The 95% confidence interval for the mean is defined by the values LL=2.0 and UL=3.4.

$CI: 2.0$

5 0

3 years ago