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3 years ago

9th grade geometry congruence proofs and reasons please no links will name brainliest will give 50 points to anyone who helps.

Mathematics

1 answer:

KengaRu [80]3 years ago

5 0

Answer: Opps- I am not in 9th grade yet- I am only in 5th grade can't help you sorry... I wish I can help you tho but i don't know what to do.

Step-by-step explanation:

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What are the values for the coefficients and constant term of 0 = 2 + 3x2 – 5x?

Mekhanik [1.2K]

This is a quadratic in the form of ax squared +bx +c

a = 3

b = -5

c = 2

Hope this helps!

3 0

3 years ago

Read 2 more answers

Pls help will give brainlist

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Answer:

WELL I DONT SEE THE PICTUE BUT WHATS THE PROBLEM?

Step-by-step explanation:

7 0

3 years ago

4 3/10 divided by 3/5

inysia [295]

Answer:

7.16

Step-by-step explanation:

5 0

4 years ago

Prove that n is odd if and only if 3n + 6 is odd by contradiction.

Reil [10]

Step-by-step explanation:

o - odd number

e - even number

n + e =o, if n is odd...rule 1

n + e = e, if n is even...rule 2

n × o = o, if n is odd...rule 3

n × o = e, if n is even...rule 4

thus if 3n +6 = o,

then 3n must be odd, following rule 1

=> 3n = o, then n is an odd number, following rule 3

4 0

3 years ago

Kim wants to determine a 90 percent confidence interval for the true proportion of high school students in the area who attend t

PolarNik [594]

Answer:

We need a sample of at least 752 students.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error of the interval is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a pvalue of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

How large of a sample must she have to get a margin of error less than 0.03

We need a sample of at least n students.

n is found when M = 0.03.

We have no information about the true proportion, so we use $\pi = 0.5$ .

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.645\sqrt{\frac{0.5*0.5}{n}}$

$0.03\sqrt{n} = 1.645*0.5$

$\sqrt{n} = \frac{1.645*0.5}{0.03}$

$(\sqrt{n})^{2} = (\frac{1.645*0.5}{0.03})^{2}$

$n = 751.67$

Rounding up

We need a sample of at least 752 students.

6 0

4 years ago