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djverab [1.8K]
3 years ago
10

Algebra help! Please help me!

Mathematics
1 answer:
aniked [119]3 years ago
6 0
Your answer would most likely be C. 2^1/3 i'm sorry if I got it wrong
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10 points!
sasho [114]
I believe it's the 3rd one. It's a really tricky question sorry if you get it wrong
4 0
3 years ago
Determine the vertex- f( x)= x^2 +4x +3<br><br> (2,-1)<br> (-2, -1)<br> (-3, -1)<br> (-1, -2)
Grace [21]

The answer is ( -2 , -1 ) .

Hope it's helped ♥️♥️♥️♥️♥️.

8 0
3 years ago
Each school bus going on the field trip holds 36 students and for adults there are six Fields buses on the field trip how many p
BartSMP [9]

Answer:

240 people

Step-by-step explanation:

According to the given data,

In each bus,there are total number of 36 students and 4 adults

So, total 40 people are in one bus so...

As, there are six Fields buses

=>40 x 6

=240

Answer: Total 240 people are going on the field trip.

5 0
4 years ago
What are 9 ten thousands
skad [1K]
This is 90,000 is your answer
8 0
3 years ago
Read 2 more answers
When Nayeli goes bowling, her scores are normally
Ber [7]

Using the normal distribution, it is found that she scores less than 128 in 28.1% of her games.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

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

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

In this problem, the mean and the standard deviation are given, respectively, by \mu = 135, \sigma = 12.

The proportion of games in which she scores less than 128 is the <u>p-value of Z when X = 128</u>, hence:

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

Z = \frac{128 - 135}{12}

Z = -0.58

Z = -0.58 has a p-value of 0.281.

She scores less than 128 in 28.1% of her games.

More can be learned about the normal distribution at brainly.com/question/24663213

#SPJ1

3 0
2 years ago
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