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Schach [20]
3 years ago
12

Solve _+_+_=30 using odd numbers​

Mathematics
1 answer:
kicyunya [14]3 years ago
3 0

Answer:

Step-by-step explanation:

15+13+3=30

Hope this helps! And Merry Christmas!

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Write the equation x2+24x+60=0 in the form (x+p)2=q
navik [9.2K]

Answer:

(x+12)^2=84

Step-by-step explanation:

x^2+24x+60=0

x^2+24x=-60

x^2+24x+(24/2)^2=-60+(24/2)^2

(x+24/2)^2=84

(x+12)^2=84 (ans)

4 0
2 years ago
24 athletes threw a shot put at a track-and-field meet. If this was 15/100 of all the athletes at the meet, how many athletes in
Andrej [43]
There were 160 athletes at the meet.
YOU'RE WELCOME :D
5 0
3 years ago
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Coordinate pairs for 12x-3y<12
timurjin [86]

Answer:

y>4x-4

Step-by-step explanation:

7 0
2 years ago
At a new exhibit in the Museum of Science, people are asked to choose between 94 or 220 random draws from a machine. The machine
Tresset [83]

Answer:

0.0869 = 8.69% probability of getting more than 61% green balls.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

The machine is known to have 99 green balls and 78 red balls.

This means that p = \frac{99}{99+78} = 0.5593

Mean and standard deviation:

\mu = p = 0.5593

s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.5593*0.4407}{99+78}} = 0.0373

a. Calculate the probability of getting more than 61% green balls.

This is 1 subtracted by the pvalue of Z when X = 0.61. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{0.61 - 0.5593}{0.0373}

Z = 1.36

Z = 1.36 has a pvalue of 0.9131

1 - 0.9131 = 0.0869

0.0869 = 8.69% probability of getting more than 61% green balls.

8 0
3 years ago
Lena says that and are like terms. Is she correct? Why or why not? A. No, because the same variables are not raised to the same
qwelly [4]

Answer:

D because of the question

8 0
3 years ago
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