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3 years ago

After substituting, what is the first step when evaluating StartFraction 4 y + 9 over 7 EndFraction when y = 8?

Mathematics

2 answers:

zaharov [31]3 years ago

5 0

Answer:

sorry if im wronggggg

Step-by-step explanation:

4y + 9

4(8) + 9

4 x 8 = 32

32 + 9 = 41

41 divided by 7 = 5.8 or 6

leonid [27]3 years ago

4 0

Answer: B.

Multiply 8 by 4

Step-by-step explanation:

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I need help (–

ExtremeBDS [4]

Answer:

Step-by-step explanation:

–

q+6)=

–

5q–30=

–

Add -5 to both sides

Subtract -5 from both sides

Multiply both sides by -5

Divide both sides by -5

Apply the distributive property

5q=

Add 30 to both sides

Divide both sides by 5 would be 69

6 0

3 years ago

Partial quotient for 654 divided by 3

Vladimir79 [104]

654 over 3 = divide the both 654 and 3 by 3.

It will give you 218 over 1.

So the answer is 218.

6 0

3 years ago

5. Three times the width of a certain rectangle exceeds twice its length by two inches. Four times its length is twelve more

damaskus [11]

Answer: A

Step-by-step explanation:

L=Length W=Width

3W=2L+2

4L=2L+2W+12

2L=2W+12

Option A

Hope this helps!! :)

Please let me know if you have any question or need further explanation

4 0

3 years ago

1. y = (x - 5)3<br><br>what are the zeros and multiplicity

Elden [556K]

Answer:

$\frac{5}{3}$ =x

Step-by-step explanation:

0=(x-5)3

0=(3x-5)

5=3x

$\frac{5}{3}$ =x

3 0

4 years ago

The mean cost of a five pound bag of shrimp is 40 dollars with a standard deviation of 7 dollars. If a sample of 51 bags of shri

eimsori [14]

Answer:

$P(|\bar{x}-40| > 0.6)=0.5404$

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 40 dollars

Standard Deviation, σ = 7 dollars

Sample size,n = 51

We are given that the distribution of cost of shrimp is a bell shaped distribution that is a normal distribution.

Formula:

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

Standard error due to sampling =

$=\dfrac{\sigma}{\sqrt{n}} = \dfrac{7}{\sqrt{51}} = 0.9801$

P(sample mean would differ by true mean by more than 0.6)

$P(|\bar{x}-40| > 0.6)\\\Rightarrow = 1-P(39.4$

$P(39.4 \leq x \leq 40.6) \\\\= P(\displaystyle\frac{39.4 - 40}{0.9801} \leq z \leq \displaystyle\frac{40.6-40}{0.9801})\\\\ = P(-0.6121 \leq z \leq 0.6121)\\= P(z \leq 0.6121) - P(z < -0.6121)\\= 0.7298 - 0.2702 =0.4596$

$P(|\bar{x}|-40 > 0.6)\\= 1-0.4596=0.5404$

0.5404 is the required probability.

4 0

3 years ago