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IceJOKER [234]
3 years ago
12

Please Help Thanks so much for the help

Mathematics
2 answers:
boyakko [2]3 years ago
7 0

Answer:

The answer is 10

Step-by-step explanation:

Airida [17]3 years ago
4 0
ANSWER: 10

REASONING: 400,000,000+7+300=400,000,307
so if you add ten you get 400,000,317
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PLEASE HELP PICTURE IS SHOWN
Alik [6]
Use the pythagorean theorem. You should come out with C. Hope this helps!
6 0
3 years ago
What is the ratio between 5 footballs and 9 baseballs
____ [38]

Answer:

5:9

Step-by-step explanation:

It's just a way to set up ratios. Whichever number is said first gets put down first, then the number after gets put after.

So, 5:9

Or, you could write it as 5 footballs to 9 baseballs.

This is just my simple understanding of what you're asking since you asked what the ratio is.

Hope I helped!

7 0
3 years ago
Heights of male students in the eighth grade are normally distributed with a mean of 57.2 in. and standard deviation of 2.4 in.
romanna [79]
<span>In statistics finding percentiles relates to the standard deviation and something called a z-score. For normally distributed data the z-score represents how many standard deviations above or below the mean that group is a part of. The z-score for normally distributed data for the 90th percentile is 1.28. The standard deviation is then multiplied by the z-score to find, in this case, the shotlrtest height needed to be in the 90th percentile of this population. In this case to be in the 90th percentile your height must be 60.27 inches.</span>
3 0
3 years ago
I NEED HELP ASAP!!!!!
viva [34]

Answer:

1 is 12

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3 0
3 years ago
Prove 2√(x) + 1/√(x + 1) &lt;= 2√(x+1) for all x in [0,inf)
EleoNora [17]
Start by multiplying each side of the inequality by \sqrt{x + 1} and simplifying:

2 \sqrt x + \frac{1}{\sqrt{x+1}}  \leq  2 \sqrt{x + 1}
(2 \sqrt x + \frac{1}{\sqrt{x+1}})(\sqrt{x + 1}) \leq (2 \sqrt{x + 1})(\sqrt{x + 1})
2 \sqrt{x(x + 1)} + 1 \leq 2(x + 1)
2 \sqrt{x^2 + x} + 1 \leq 2x + 2
2 \sqrt{x^2 + x} \leq 2x + 1
\sqrt{x^2 + x} \leq x + \frac{1}{2}

From here, we can square both sides to get

x^2 + x \leq (x + \frac{1}{2})^2
x^2 + x \leq x^2 + x + \frac{1}{4}
0  \leq \frac{1}{4}, which is always true.
3 0
3 years ago
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