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baherus [9]
4 years ago
15

What is the minimum value for g(x)=x2−18x+79? Enter your answer in the box.

Mathematics
1 answer:
iVinArrow [24]4 years ago
7 0
The minimum value is -2
You might be interested in
When the input is 200 what is the output? Explain please.
anygoal [31]
The slope formula is delta y/delta x=m
You can choose any two points. Let's choose (0,25) and (1,30)
30-25/1-0=5/1=5
Slope intercept form is y=mx+b
b=the y intercept
y=5x+25 since (0,25) is the coordinates for the y inthercept.
Plug in 200: y=5(200)+25
Simplify and you get that y=1025
Your answer is 1025
Your welcome
4 0
3 years ago
Calculate the length of the altitude (h) of the rhombus
allochka39001 [22]

Answer:

h = 15

Step-by-step explanation:

Use Pythagorean Theorem to solve for h.

a² + b² = c²

8² + h² = 17²

64 + h² = 289

h² = 225

h = 15

7 0
3 years ago
3x\4=2y+6xy. what is "x" equal too? what is "y" equal too?
andreyandreev [35.5K]
 the answer is 
x=8y/3(1-8y)


y=3x/8(1+3x)

hope this helped l used cymath to solve this problem. 
good day : )
6 0
3 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
What is the probability of choosing a pair that contains ONLY cards labeled with the letters B, C, or F?
Nezavi [6.7K]

Answer:

Pr = \frac{1}{6}

Step-by-step explanation:

Given

Cards = \{A,B,C,D,E,F\}

r = 2 --- selection

Required

Probability of choosing a pair with only B, C and F

The total selection is: 36

This is calculated as thus:

First = 6 --- First selection can be any of the 6 cards

Second = 6 --- Second selection can be any of the 6 cards [No restriction]

So, we have:

n = First * Second

n = 6 * 6

n = 36

The pair of only B, C or F is:

Pair = \{(B,C), (B,F), (C,B),(C,F),(F,B),(F,C)\}

n(Pair) = 6

So, the probability is:

Pr = \frac{n(Pair)}{n}

Pr = \frac{6}{36}

Simplify

Pr = \frac{1}{6}

4 0
3 years ago
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