A new family has moved in next door. You know the family has two children. What is the probability that both children are boys?

Write an exponential function in the form y=ab^x that goes through points (0,12) and (3,6144)

skad [1K]

Given:

An exponential function goes through points (0,12) and (3,6144).

To find:

The exponential function.

Solution:

The general exponential function is:

$y=ab^x$ ...(i)

The function goes through point (0,12). Substituting $x=0,y=12$ , we get

$12=ab^0$

$12=a$

The function goes through point (3,6144). Substituting $a=12, x=3,y=6144$ in the general exponential function, we get

$6144=12b^3$

$\dfrac{6144}{12}=b^3$

$512=b^3$

$512^{\frac{1}{3}}=b$

$8=b$

Putting $a=12,b=8$ in (i), we get

$y=12(8)^x$

Therefore, the required exponential function is $y=12(8)^x$ .

3 0

3 years ago

4. The Gold family sold their house for $450,000. They paid a realty

Rudiy27

Answer:

27,000

Step-by-step explanation:

$450,000 x 0.06= $27,000

0.06 is the 6% that you multiply by the total amount they earned to see how much they paid the realty company.

7 0

3 years ago

In a survey of 23312331 adults, 730730 say they believe in UFOs. Construct a 99 %99% confidence interval for the population pro

serg [7]

Answer:

0.288 < p < 0.338

Step-by-step explanation:

We have 2331 adults polled and 730 said they believe in UFOs. We are to construct a 99% confidence interval for the population proportion. We are told to round to 3 decimal places as needed.

We need to find p-hat and q-hat

p-hat = 730/2331 = 0.313

q-hat = 1 - p-hat = 1 - 0.313 = 0.687

Since the sample size is greater than 30, we use a z-value for 99% confidence, which is 2.575

See attached photo for the construction of the confidence interval

8 0

3 years ago

The manager of a supermarket would like to determine the amount of time that customers wait in a check-out line. He randomly sel

garri49 [273]

Complete question:

The manager of a supermarket would like to determine the amount of time that customers wait in a check-out line. He randomly selects 45 customers and records the amount of time from the moment they stand in the back of a line until the moment the cashier scans their first item. He calculates the mean and standard deviation of this sample to be barx = 4.2 minutes and s = 2.0 minutes. If appropriate, find a 90% confidence interval for the true mean time (in minutes) that customers at this supermarket wait in a check-out line

Answer:

(3.699, 4.701)

Step-by-step explanation:

Given:

Sample size, n = 45

Sample mean, x' = 4.2

Standard deviation $\sigma$ = 2.0

Required:

Find a 90% CI for true mean time

First find standard error using the formula:

$S.E = \frac{\sigma}{\sqrt{n}}$

$= \frac{2}{\sqrt{45}}$

$= \frac{2}{6.7082}$

$SE = 0.298$

Standard error = 0.298

Degrees of freedom, df = n - 1 = 45 - 1 = 44

To find t at 90% CI,df = 44:

Level of Significance α= 100% - 90% = 10% = 0.10

$t_\alpha_/_2_, _d_f = t_0_._0_5_, _d_f_=_4_4 = 1.6802$

Find margin of error using the formula:

M.E = S.E * t

M.E = 0.298 * 1.6802

M.E = 0.500938 ≈ 0.5009

Margin of error = 0.5009

Thus, 90% CI = sample mean ± Margin of error

Lower limit = 4.2 - 0.5009 = 3.699

Upper limit = 4.2 + 0.5009 = 4.7009 ≈ 4.701

Confidence Interval = (3.699, 4.701)

5 0

3 years ago

Find the x-coordinates of any relative extrema and inflection point(s) for the function f(x) = 6x(1/3) + 3x(4/3). You must justi

stealth61 [152]

Applying our power rule gets us our first derivative,

$\rm f'(x)=6\frac13x^{-2/3}+3\cdot\frac43x^{1/3}$

simplifying a little bit,

$\rm f'(x)=2x^{-2/3}+4x^{1/3}$

looking for critical points,

$\rm 0=2x^{-2/3}+4x^{1/3}$

We can apply more factoring.
I hope this next step isn't too confusing.
We want to factor out the smallest power of x from both terms,
and also the 2 from each.

$0=2x^{-2/3}\left(1+2x\right)$

When you divide x^(-2/3) out of x^(1/3),
it leaves you with x^(3/3) or simply x.

Then apply your Zero-Factor Property,

$\rm 0=2x^{-2/3}\qquad\qquad\qquad 0=(1+2x)$

and solve for x in each case to find your critical points.

Apply your First Derivative Test to further classify these points. You should end up finding that x=-1/2 is an relative extreme value, while x=0 is not.

Let's come back to this,

$\rm f'(x)=2x^{-2/3}+4x^{1/3}$

and take our second derivative.

$\rm f''(x)=-\frac43x^{-5/3}+\frac43x^{-2/3}$

Looking for inflection points,

$\rm 0=-\frac43x^{-5/3}+\frac43x^{-2/3}$

Again, pulling out the smaller power of x, and fractional part,

$\rm 0=-\frac43x^{-5/3}\left(1-x\right)$

And again, apply your Zero-Factor Property, setting each factor to zero and solving for x in each case. You should find that x=0 and x=1 are possible inflection points.

Applying your Second Derivative Test should verify that both points are in fact inflection points, locations where the function changes concavity.

8 0

3 years ago