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2 years ago

Find the coordinates of the midpoint of LB if L(8,5) and B(-6,2) 10 points!!!

Mathematics

1 answer:

KATRIN_1 [288]2 years ago

3 0

Answer:

A, (1, 3+1/2)

Step-by-step explanation:

Midpoint formula for reference: m= {(x1 + x2)/2, (y1 + y2)/2}

Plugging in the points we get: m= {(8 - 6)/2, (5 + 2)/2}

Now we simplify using PEMDAS. First step is parentheses.

m= {2/2, 7/2}

Simplifying again (and making 7/2 a mixed number), it becomes

m= {1, 3+1/2}

Hope this helps!

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3 years ago

Which methods correctly solve for the variable s in the equation -2 - 5s =9?

Ivanshal [37]

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2 years ago

Which of the following is the conjugate of a complex number with 2 as the real part and −8 as the imaginary part?

vampirchik [111]

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3 years ago

Read 2 more answers

Jack had 3 bottles of soda. Each bottle held 2 1/2 cups of soda. How many cups of soda did he have total?

lukranit [14]

Answer:

1 1/5 cups of soda.

Step-by-step explanation:

To solve this you need to change 3 and 2 1/2 into improper fractions so you get 6/2 and 5/2. Then you have to flip 5/2 so it becomes 2/5 and then you just have to cross cancel. Since the 2 from 2/5 can go into the 2 in 6/2 one time, 2/5 changes into 1/5 and 6/2 changes to 6/1. Now you can multiply across so 6 times 1 is 6 and 1 times 5 is 5 so you get 6/5. Now you have to change it into a mixed number which would be 1 1/5.

I hope this helps you :D

7 0

3 years ago

A chemist examines 12 sedimentary samples for bromide concentraction. The mean bromide concentration for the sample date is 0.43

Umnica [9.8K]

Answer:

a) $z = 1.645$

b) Lower endpoint: 0.422cc/m³

Upper endpoint: 0.452 cc/m³

Step-by-step explanation:

Population is approximately normal, so we can find the normal confidence interval.

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.05 = 0.95$ , so $z = 1.645$ . This is the critical value, the answer for a).

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.645*\frac{0.0325}{\sqrt{12}} = 0.015$

The lower end of the interval is the sample mean subtracted by M. So it is 0.437 - 0.015 = 0.422cc/m³.

The upper end of the interval is the sample mean added to M. So it is 0.437 + 0.015 = 0.452 cc/m³.

Lower endpoint: 0.422cc/m³

Upper endpoint: 0.452 cc/m³

4 0

3 years ago