The function f(x) is shown in the graph.<br><br> What is the equation for f(x)?

Determine whether y varies with x. If so, find the constant of variation and write the function rule.

ira [324]

Answer: Yes, y does vary directly with x.

Constant of variation = 1/4

The function rule is y = (1/4)x

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Explanation:

Let's assume that y does vary directly with x.

If that's the case, then we have an equation in the form y = kx, where k is the constant of variation.

Solving for k gets us k = y/x

For each row, divide the y value over the x value

row one: k = y/x = 14/56 = 0.25
row two: k = y/x = 20/80 = 0.25
row three: k = y/x = 22/88 = 0.25

Each row yields the value k = 0.25 and it fully confirms y does vary directly with x.

So y = kx becomes y = 0.25x as the function rule, which is equivalent to y = (1/4)x

6 0

1 year ago

A college basketball coach has 12 players on his team. eight players are receiving scholarships, and four are not. the coach dec

EleoNora [17]

Solution: The given problem is a binomial distribution.

The probability that a team member receives a scholarship is $\frac{8}{12} = 0.67$

Therefore, the given problem follows binomial with n = 5 and p = 0.6667

Now the probability that 4 of 5 players selected are on scholarship is:

$P(x=4) = \binom{5}{4} \times 0.6667^{4} (1-0.6667)^{5-4}$

$=5 \times 0.1976 \times 0.3333$

$=0.329$

Therefore, the probability that 4 of 5 players selected are on scholarship is 0.329

5 0

3 years ago

kendrick deposits twice as much money into his account as he did the day before his initial deposit is 5 dollars write an equati

Ulleksa [173]

To represent this scenario, you will look at what is happening mathematically and represent it.

Beginning with 5, this will double each day, so 5 x 2 is the deposit.

Then, this amount will double, so (5 x 2) x 2 is the deposit.

Next this will double, and so on.

What is happening is that it is always doubling the previous day's deposit, so x 2 x 2 x 2, etc... you will use exponents to show the number of days that it doubled.

y = 5 x 2^x

8 0

3 years ago

I need help please, i dont understand

jenyasd209 [6]

9514 1404 393

Answer:

(a) none of the above

Step-by-step explanation:

The largest exponent in the function shown is 2. That makes it a 2nd-degree function, also called a quadratic function. The graph of such a function is a parabola -- a U-shaped curve.

The coefficient of the highest-degree term is the "leading coefficient." In this case, that is the coefficient of the x² term, which is 1. When the leading coefficient of an even-degree function is positive, the U curve has its open end at the top of the graph. We say it "opens upward." (When the leading coefficient is negative, the curve opens downward.)

This means the bottom of the U is the minimum value the function has. For a quadratic in the form ax²+bx+c, the horizontal location of the minimum on the graph is at x=-b/(2a). This extreme point on the curve is called the "vertex."

This function has a=1, b=1, and c=3. The minimum of the function is where ...

x = -b/(2·a) = -1/(2·1) = -1/2

This value is not listed among the answer choices, so the correct choice for this function is ...

none of the above

__

The attached graph of the function confirms that the minimum is located at x=-1/2

_____

<em>Additional comment</em>

When you're studying quadratic functions, there are few formulas that you might want to keep handy. The formula for the location of the vertex is one of them.

8 0

2 years ago

Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o

kirza4 [7]

Answer:

a) H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

b) $df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

c) $t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

$\sigma_o =1.34$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic (chi square test)

$\alpha=0.01$ significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

The statistic is given by:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

Part b

The degrees of freedom are given by:

$df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

Part c

Replacing the info we got:

$t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

5 0

2 years ago

The function f(x) is shown in the graph. What is the equation for f(x)?