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3 years ago

PLEASE HELP ME YOU GUYSSS

Mathematics

1 answer:

notsponge [240]3 years ago

3 0

Answer:

c i got it right on test

Step-by-step explanation:

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Solve it using Pythagoras theorem <br>plzz help I will mark u brainliest

mrs_skeptik [129]

Answer:

Reasoning:

(7^2)-(3^2)= (6.32^2)

(6.32^2)+(6^2)=(8.72^2)

I hope this helps and good luck

6 0

3 years ago

The nth term of 50, 80, 110 is...

Olenka [21]

Answer:

$a_n=50+(n-1)30$

Step-by-step explanation:

Given series is,

50,80,110....

First term, a = 50

Common difference, d = 80-50 = 30

We need to find the nth term of the given sequence.

The nth term of an AP is given by :

$a_n=a+(n-1)d$

Put a = 50 and d = 30 in the above formula

$a_n=50+(n-1)30$

Hence, the nth term of the sequence is $a_n=50+(n-1)30$ .

5 0

3 years ago

Some investments in the stock market have earned 10% annually. At this rate, earning can be found using the formula A=p(1.10)n,

Harlamova29_29 [7]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
The answer to the above question is C which is $8,339.88. below is the solution:

just plug in 1500 for p, and 18 for n:

<span>1500(1.10)^18 = 8339.88</span>

4 0

3 years ago

Can anyone please help me with this?

malfutka [58]

X^2(x+5)+9(x+5)

(x^2+9)(x+5)

8 0

3 years ago

You are trying to estimate the average amount a family spends on food during a year. In the past the standard deviation of the a

svetoff [14.1K]

Answer:

$n=(\frac{2.58(1000)}{50})^2 =2662.56 \approx 2663$

So the answer for this case would be n=2663 rounded up to the nearest integer

Step-by-step explanation:

We have the following info:

$ME = 50$ margin of error desired

$\sigma = 1000$ the standard deviation for this case

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =50 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 99% of confidence interval now can be founded using the normal distribution. The significance is $\alpha=0.01$ . And for this case would be $z_{\alpha/2}=2.58$ , replacing into formula (b) we got:

$n=(\frac{2.58(1000)}{50})^2 =2662.56 \approx 2663$

So the answer for this case would be n=2663 rounded up to the nearest integer

5 0

3 years ago