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vitfil [10]
3 years ago
12

THIS IS DUE TODAY PLS PLS PLS HELP ME ON IT :(

Mathematics
2 answers:
Rufina [12.5K]3 years ago
7 0

Answer:

monke

Step-by-step explanation:

monkey

AnnyKZ [126]3 years ago
7 0

I think its the first and last one. But I could be completely wrong.

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For the standard form of two billion three Hundred fifty thousand four Danielle wrote 2,350,400,000 what error did she make what
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I believe the correct answer is 2,300,500,004
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Simple question, 4th.. Please help :( Thank you for the help so far! :D
aliya0001 [1]
I'm pretty sure the answer to your question would be A
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Evaluate. 7^2 - 3 + 9 × 8 ÷ 2<br><br> w =
julia-pushkina [17]
7^2 - 3 + 9 x 8 / 2
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3 0
3 years ago
Read 2 more answers
A professor at a local university noted that the exam grades of her students were normally distributed with a mean of 73 and a s
jasenka [17]

Answer:

The minimum score of those who received C's is 67.39.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 73, \sigma = 11

If 69.5 percent of the students received grades of C or better, what is the minimum score of those who received C's?

This is X when Z has a pvalue of 1-0.695 = 0.305. So it is X when Z = -0.51.

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

-0.51 = \frac{X - 73}{11}

X - 73 = -0.51*11

X = 67.39

The minimum score of those who received C's is 67.39.

7 0
3 years ago
Help please I need 5 points total. i need 2 to left of vertex. i need vertex. i need 2 to right of vertex. Please, quickly, I am
zlopas [31]

We are given with a quadratic equation which represents a Parabola , we need to find the vertex of the parabola , But let's recall that , For any quadratic equation of the form ax² + bx + c = 0 , the vertex of the parabola is given by ;

{\quad \qquad \boxed{\bf{Vertex = \left(\dfrac{-b}{2a},\dfrac{-D}{4a}\right)}}}

Where , D = b² - 4ac (Discriminant)

Now , On comparing the given equation with ax² + bx + c , we have

⇢⇢⇢ <em><u>a = 1 , b = - 10 , c = 2</u></em><em><u>7</u></em>

Now , Calculating D ;

{:\implies \quad \sf D=(-10)^{2}-4\times 1\times 27}

{:\implies \quad \sf D=100-108}

{:\implies \quad \bf \therefore \quad D=-8}

Now , Calculating the vertex ;

{:\implies \quad \sf Vertex =\bigg\{\dfrac{-(-10)}{2\times 1},\dfrac{-(-8)}{4\times 1}\bigg\}}

{:\implies \quad \sf Vertex = \left(\dfrac{10}{2},\dfrac{8}{2}\right)}

{:\implies \quad \bf \therefore \quad Vertex = (5,2)}

Hence , The vertex of the parabola is at (5,2)

Note :- As the Discriminant < 0 . So , the equation will have imaginary roots .

Refer to the attachment for the graph as well .

5 0
2 years ago
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