3 years ago

Solve and simplify 3. (y^2-7y-5)-(-3y^2+3y-4)

Mathematics

1 answer:

yKpoI14uk [10]3 years ago

7 0

$(y^2-7y-5)-(-3y^2+3y-4)=y^2-7y-5-(-3y^2)-3y-(-4)\\\\=y^2-7y-5+3y^2-3y+4$

<em>combine like terms</em>

$=(y^2+3y^2)+(-7y-3y)+(-5+4)=\boxed{4y^2-10y-1}$

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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

Umnica [9.8K]

Answer:

88.51 is the minimum score needed to receive a grade of A.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 73

Standard Deviation, σ = 11

We are given that the distribution of exam grades is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.0793.

$P( X \geq x) = P( z \geq \displaystyle\frac{x - 73}{11})=0.0793$

$= 1 -P( z \leq \displaystyle\frac{x - 73}{11})=0.0793$

$=P( z \leq \displaystyle\frac{x - 73}{11})=0.9207$

Calculation the value from standard normal z table, we have,

$P(z \leq 1.410) = 0.9207$

$\displaystyle\frac{x - 73}{11} = 1.410\\x =88.51$

Hence, 88.51 is the minimum score needed to receive a grade of A.

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12% of 72 is what number?

xxTIMURxx [149]

I hope this helps you ?.12%=72 ?.12=72.100 ?=6.100 ?=600

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Answer:

Number 7: 22.10 divided by 2.6 = 8.5 or 8.50 (depending on what your teacher prefers.

Step-by-step explanation:

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