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2 years ago

These tables of value represent continuous functions. In which table do the values represent an exponential function?

Mathematics

1 answer:

oksano4ka [1.4K]2 years ago

5 0

Answer:

Step-by-step explanation:

A goes up by 1

C goes up by 8

D goes up by 5

But B increments by 2x (2 times)

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A recent survey in the N.Y. Times Almanac indicated that 48.8% of families own stock. A broker wanted to determine if this surve

nasty-shy [4]

Answer:

There is enough statistical evidence to support the claim that the survey is not accurate.

Step-by-step explanation:

We have to perform a test of hypothesis on the proportion.

The claim is that the proportion of families that own stock differs from 48.8%.

Then, the null and alternative hypothesis are:

$H_0: \pi=0.488\\\\H_a:\pi\neq0.488$

The significance level is 0.05.

The sample. of size n=250, has a proportion of p=0.568.

$p=X/n=142/250=0.568$

The standard error of the proportion is

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.488*0.512}{250}}=\sqrt{0.00099}=0.032$

The z-statistic can now be calculated:

$z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.568-0.488-0.5/250}{0.032}=\dfrac{0.078}{0.032}=2.4375$

The P-value for this two-tailed test is then:

$P-value=2*P(z>2.4375)=0.015$

As the P-value is smaller than the significance level, the effect is significant. The null hypothesis is rejected.

There is enough evidence to support the claim that the survey is not accurate.

6 0

2 years ago

HELP!!!

NikAS [45]

Answer:

I believe it’s the 3rd one.

Explanation:

It’s the only graph where it looks like (2.5,5) was graphed.

7 0

2 years ago

Read 2 more answers

We want to estimate the population mean within 5, with a 99% level of confidence. the population standard deviation is estimated

pashok25 [27]

A sample size of 60 is required.

We use the formula
$n=(\frac{z*\sigma}{E})^2$

We first find the z-score associated with this level of confidence:
Convert 99% to a decimal: 99/100 = 0.99
Subtract from 1: 1-0.99 = 0.01
Divide by 2: 0.01/2 = 0.005
Subtract from 1: 1-0.005 = 0.995

Using a z-table (http://www.z-table.com) we see that this value is equally distant from 2.57 and 2.58; therefore we will use 2.575:

$n=(\frac{15(2.575)}{5})^2=59.68\approx60$

3 0

2 years ago

A large tank is partially filled with 100 gallons of fluid in which 20 pounds of salt is dissolved. Brine containing 1 2 pound o

Valentin [98]

Answer:

47.25 pounds

Step-by-step explanation:

$\dfrac{dA}{dt}=R_{in}-R_{out}$

<u>First, we determine the Rate In</u>

Rate In=(concentration of salt in inflow)(input rate of brine)

$=(0.5\frac{lbs}{gal})( 6\frac{gal}{min})\\R_{in}=3\frac{lbs}{min}$

Change In Volume of the tank, $\frac{dV}{dt}=6\frac{gal}{min}-4\frac{gal}{min}=2\frac{gal}{min}$

Therefore, after t minutes, the volume of fluid in the tank will be: 100+2t

Rate Out=(concentration of salt in outflow)(output rate of brine)

$R_{out}=(\frac{A(t)}{100+2t})( 4\frac{gal}{min})\\\\R_{out}=\frac{4A(t)}{100+2t}$

Therefore:

$\dfrac{dA}{dt}=3-\dfrac{4A(t)}{100+2t}\\\\\dfrac{dA}{dt}=3-\dfrac{4A(t)}{2(50+t)}\\\\\dfrac{dA}{dt}=3-\dfrac{2A(t)}{50+t}\\\\\dfrac{dA}{dt}+\dfrac{2A(t)}{50+t}=3$

This is a linear differential equation in standard form, therefore the integrating factor:

$e^{\int \frac{2}{50+t}dt}=e^{2\ln|50+t|}=e^{\ln(50+t)^2}=(50+t)^2$

Multiplying the DE by the integrating factor, we have:

$(50+t)^2\dfrac{dA}{dt}+(50+t)^2\dfrac{2A(t)}{50+t}=3(50+t)^2\\\{(50+t)^2A(t)\}'=3(50+t)^2\\$Taking the integral of both sides\\\int \{(50+t)^2A(t)\}'= \int 3(50+t)^2\\(50+t)^2A(t)=(50+t)^3+C $ (C a constant of integration)\\Therefore:\\A(t)=(50+t)+C(50+t)^{-2}$