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drek231 [11]
3 years ago
6

Based on the family the graph below belongs to, which equation could represent the graph? On a coordinate plane, 2 curves are sh

own. One curve approaches x = negative 2 in quadrant 2 and then decreases into quadrant 1. It crosses the y-axis at (0, 3.5). Another curve approaches x = negative 2 in quadrant 3 and then increases into quadrant 2. It crosses the x-axis at (negative 2.5, 0). y = StartFraction 1 Over x + 2 EndFraction + 3 y = 0.2 Superscript x Baseline + 3 y = x cubed + 3 y = log x + 3
Mathematics
2 answers:
Greeley [361]3 years ago
5 0

Answer:

a. y = 1/(x + 2) + 3

Step-by-step explanation:

From the description of the function, it is seen that the curve has an asymptote at x = -2. Asymptotes are typical from rational functions. The only option that matches this feature is y = 1/(x + 2) + 3

Moreover, replacing x = 0 into the equation, we get:

y = 1/(0 + 2) + 3 = 3.5

And replacing y = 0 into the equation, we get:

0 = 1/(x + 2) + 3

-3 = 1/(x + 2)

x + 2 = -1/3

x = -1/3 - 2

x = -2.33 ≈ -2.5

Rainbow [258]3 years ago
4 0

Answer:

its A

Step-by-step explanation:

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leva [86]

Answer:

120x + 245 = y

x = number of additional tiers

(edited to be in the proper order)

8 0
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7. (6-4)3 pls help need answer quick pls
Dennis_Churaev [7]

Answer:

6

Step-by-step explanation:

(6-4)3

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Suppose a student's score on a standardize test to be a continuous random variable whose distribution follows the Normal curve.
blsea [12.9K]

Answer:

a) Percentage of students scored below 300 is 1.79%.

b) Score puts someone in the 90th percentile is 638.

Step-by-step explanation:

Given : Suppose a student's score on a standardize test to be a continuous random variable whose distribution follows the Normal curve.

(a) If the average test score is 510 with a standard deviation of 100 points.

To find : What percentage of students scored below 300 ?

Solution :

Mean \mu=510,

Standard deviation \sigma=100

Sample mean x=300

Percentage of students scored below 300 is given by,

P(Z\leq \frac{x-\mu}{\sigma})\times 100

=P(Z\leq \frac{300-510}{100})\times 100

=P(Z\leq \frac{-210}{100})\times 100

=P(Z\leq-2.1)\times 100

=0.0179\times 100

=1.79\%

Percentage of students scored below 300 is 1.79%.

(b) What score puts someone in the 90th percentile?

90th percentile is such that,

P(x\leq t)=0.90

Now, P(\frac{x-\mu}{\sigma} < \frac{t-\mu}{\sigma})=0.90

P(Z< \frac{t-\mu}{\sigma})=0.90

\frac{t-\mu}{\sigma}=1.28

\frac{t-510}{100}=1.28

t-510=128

t=128+510

t=638

Score puts someone in the 90th percentile is 638.

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3 years ago
Find the value of x.<br> (3x + 10)<br> 98°
Aleks04 [339]

Answer:

-3

Step-by-step explanation:

(3x + 10)98°

294x+980

294x=-980

X=-980/294

X=-3

So the value of X is -3

5 0
3 years ago
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