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3 years ago

In which direction does the right side of this graph point? F(x)=3x^4-x^2+4x-2

Mathematics

2 answers:

kaheart [24]3 years ago

4 0

Answer:

Positive direction (Up) $\uparrow$

Step-by-step explanation:

The given function is

$f(x)=3x^4-x^2+4x-2$

The degree of this polynomial is even.

The leading coefficient is positive 3.

The polynomial will rise $\uparrow$ on the left and rise $\uparrow$ on the right.

Therefore the right side of the graph of $f(x)=3x^4-x^2+4x-2$ points in the positive direction of the y-axis.

Sonja [21]3 years ago

4 0

Answer:

Step-by-step explanation:

Apex

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You are running a concession stand at a basketball game. You are selling hotdogs and sodas. Each hotdog cost $1.50 and each soda

Lubov Fominskaja [6]

Answer:

35 hotdogs

Step-by-step explanation:

You are running a concession stand at a basketball game. You are selling hotdogs and sodas.

Let the number of hot dogs be represented by x

The number of soda be represented by y

You sold a total of 87 hotdogs and sodas combined

x + y = 87

x = 87 - y

Each hotdog cost $1.50 and each soda cost $0.50. At the end of the night you made a total of $78.50.

Hence we have the equation:

$1.50 × x + $0.50 × y = $78.50

1.50x + 0.50y = 78.50

Substitute 87 - y for x

1.50(87 - y) + 0.50y = 78.50

130.5 - 1.50y + 0.50y = 78.50

Collect like terms

- 1.50y + 0.50y = 78.50 - 130.50

-1.00y = -52

y = -52/-1

y = 52 sodas

How many hotdogs did you sell?

Using the equation:

x = 87 - y

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3 years ago

(9, -4); slope =-2/3 PLEASE HELP

kupik [55]

Answer: next point would be (12,-6)

Step-by-step explanation:

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3 years ago

The service life of a battery used in a cardiac pacemaker is assumed to be normally distributed. A random sample of ten batterie

Fittoniya [83]

Answer:

The confidence interval is $25.16 < \mu < 26.85$

Step-by-step explanation:

From the question we are given a data set

25.5, 26.1, 26.8, 23.2, 24.2, 28.4, 25.0, 27.8, 27.3, and 25.7.

The mean of the this sample data is

$\= x = \frac{\sum x_i}{n}$

where is the sample size with values n = 10

$\= x = \frac{25.5+ 26.1+ 26.8+23.2+ 24.2+ 28.4+ 25.0+ 27.8+ 27.3+ 25.7}{10}$

$\= x = 26$

The standard deviation is evaluated as

$\sigma = \sqrt{\frac{\sum (x-\= x)}{n} }$

substituting values

$= \sqrt{\frac{ ( 25.5-26)^2, (26.1-26)^2, (26.8-26)^2, (23.2-26)^2}{10} }$

$\cdot \ \cdot \ \cdot \sqrt{\frac{ ( 24.2-26)^2, (28.4-26)^2+( 25.0-26)^2+ (27.8-26)^2+( 27.3-26)^2+( 25.7-26)^2}{10} }$

$\sigma = 1.625$

The confidence level is given as 90% hence the level of significance is calculated as

$\alpha = 100 -90$

$\alpha =10$ %

$\alpha = 0.10$

Now the critical values of $\frac{\alpha }{2}$ is obtained from the normal distribution table as

$Z_{\frac{\alpha }{2} } = 1.645$

The reason we are obtaining the critical values of $\frac{\alpha }{2}$ instead of $\alpha$ is because we are considering two tails of the area under the normal curve

The margin of error is evaluated as

$MOE = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

substituting values

$MOE = 1.645 * \frac{1.625 }{\sqrt{10} }$

$MOE = 0.845$

The 90%, two sided confidence interval is mathematically evaluated as

$\= x - MOE < \mu < \= x + MOE$

$26 - 0.845 < \mu < 26 + 0.845$

$25.16 < \mu < 26.85$

Given that the lower and the upper limit is greater than 25 then we can assure the manufactures that the battery life exceeds 25 hours

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