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2 years ago

The graph below could be the graph of which exponential function?

Mathematics

2 answers:

GrogVix [38]2 years ago

8 0

Answer:

The graph could be a graph of the function:

Option: B B. $F(x)=2\cdot (0.5)^x$

Step-by-step explanation:

We know that for any exponential function of the type:

$f(x)=ab^x$

if b>1 then it represent a exponential growth

and if 0<b<1 then it represent a exponential decay.

Clearly from the graph given to us we could observe that it is a graph of a exponential decay.

Hence 0<b<1

Now we will check for each of the given options:

$F(x)=2\cdot (1.6)^x$

Now as b=1.6>1

So, this function will represent a exponential growth.

Hence, option: A is incorrect.

$F(x)=2\cdot (7)^x$

Now as b=7>1

So, this function will represent a exponential growth.

Hence, option: C is incorrect.

$F(x)=2^x$

Now as b=2>1

So, this function will represent a exponential growth.

Hence, option: D is incorrect.

$F(x)=2\cdot (0.5)^x$

Now as b=0.5<1 and also 0.5>0

So, this function will represent a exponential decay

Hence, option: B is correct.

MrMuchimi2 years ago

7 0

Your answer is B F(x) * (0.5)^x

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23.6 (approximate value)

Step-by-step explanation:

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2 years ago

A wheat farmer cuts down the stalks of wheat and gathers them in 200 piles. The 200 gathered piles will be put on a truck. The t

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Answer:

Check the explanation

Step-by-step explanation:

We want to estimate the total weight of grain on the field based on the data on a simple random sample of 5 piles out of 200. The population and sample sizes are N=200 & n=5 respectively.

1) Let Y_1,Y_2,...,Y_{200} be the weight of grain in the 200 piles and y_1,y_2,...,y_{5} be the weights of grain in the pile from the simple random sample.

We know, the sample mean is an unbiased estimator of the population mean. Therefore,

$\widehat{\mu}=\overline{y}=\frac{1}{5}\sum_{i=1}^{5}y_i=\frac{1}{5}(3.3+4.1+4.7+5.9+4.5)=4.5$

where \mu is the mean weight of grain for all the 200 piles.

Hence, the total grain weight of the population is

$\widehat{Y}=Y_1+Y_2+...+Y_{200}$

= $200\times \widehat{\mu}\: \: \: =200\times 4.5\: \: \: =900\, lbs$

2) To calculate a bound on the error of estimates, we need to find the sample standard deviation.

The sample standard deviation is

S= $\sqrt{\frac{1}{5-1}\sum_{i=1}^{5}(y_i-\overline{y})^2}\: \: \: =0.9486$

Then, the standard error of $\widehat{Y} is$

$\sigma_{\widehat{Y}} =\sqrt{\frac{N^2S^2}{n}\bigg(\frac{N-n}{N}\bigg)}\: \: \:$ =83.785

Hence, a 95% bound on the error of estimates is

$[\pm z_{0.025}\times \sigma_{\widehat{Y}}]\: \: \: =[\pm 1.96\times 83.875]\: \: \: =[\pm 164.395]$

3) Let x_1,x_2,...,x_5 denotes the total weight of the sampled piles.

Mean total weight of the sampled piles is

$\overline{x}=\frac{1}{5}\sum_{i=1}^{5}x_i=45$

The sample ratio is

$r=\frac{\overline{y}}{\overline{x}}=\frac{4.5}{45}$ =0.1 , this is also the estimate of the population ratio R= $\frac{\overline{Y}}{\overline{X}} .$

Therefore, the estimated total weight of grain in the population using ratio estimator is

$\widehat{Y}_R\: \: =r\times 8800\: \: =0.1\times 8800\: \: =880\,$ lbs

4) The variance of the ratio estimator is

var(r)= $\frac{N-n}{N}\frac{1}{n}\frac{1}{\mu_x^2}\frac{\sum_{i=1}^{5}(y_i-rx_i)^2}{n-1} , where \mu_x=8800/200=44lbs$

= $\frac{200-5}{200}\, \frac{1}{5}\: \frac{1}{44^2}\, \frac{0.2}{5-1}=0.000005$

Hence, the standard error of the estimate of the total population is

$\sigma_R=\sqrt{X^2 \: var(r)}\: \: \: =\sqrt{8800^2\times 0.000005}\: \: \:$ =21.556

Hence, a 95% bound on the error of estimates is

$[\pm z_{0.025}\times \sigma_{R}]\: \: \: =[\pm 1.96\times 21.556]\: \: \: =[\pm 42.25]$

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2 years ago

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Machines A and B always operate independently and at their respective constant rates. When working alone, Machine A can fill a p

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Answer:

The value of x is $\frac{10}{3}$ hours.

Step-by-step explanation:

Machine A = 5 hours

Machine B = x hours

Machine A and B = 2 hours

Using the formula: $\frac{T}{A} + \frac{T}{B} = 1$

where:

T is the time spend by both machine

A is the time spend by machine A

B is the time spend by machine B

$\frac{2}{5} + \frac{2}{x} = 1$

Let multiply the entire problem by the common denominator (5B)

$5x(\frac{2}{5} + \frac{2}{x} = 1)$

2x + 10 = 5x

Collect the like terms

10 = 5x - 2x

10 = 3x

3x = 10

Divide both side by the coefficient of x (3)

$\frac{3x}{3} = \frac{10}{3}$

$x = \frac{10}{3}$ hours.

Therefore, Machine B will fill the same lot in $\frac{10}{3}$ hours.

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2 years ago

Debra and Ian shared in $1,000,000 estate. If Ian received $125,000 and debra the rest, what fraction of the estate did Debra re

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Debra received 7/8
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Which statements about the local maximums and minimums for the given function are true? Choose three options.

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Answer:

Over the interval [2, 4], the local minimum is –8.

Over the interval [3, 5], the local minimum is –8.

Over the interval [1, 4], the local maximum is 0.

Step-by-step explanation:

The true statements are:

Over the interval [2, 4], the local minimum is –8.

Over the interval [3, 5], the local minimum is –8.

Over the interval [1, 4], the local maximum is 0.

Lets discuss each option one by one:

Over the interval [1, 3], the local minimum is 0

This is a false statement. Look at the graph. The minimum point given is (3.4,-8). Therefore the local minimum is -8 not 0

Over the interval [2, 4], the local minimum is –8.

This statement is true because the given minimum point is(3.4, -8). Thus the local minimum is -8 which is true

Over the interval [3, 5], the local minimum is –8.

According to the given minimum point, the local minimum is -8 which is true

Over the interval [1, 4], the local maximum is 0.

Look at the graph. The maximum point given is (2,0). Thus this statement is true because local maximum is 0.

Over the interval [3, 5], the local maximum is 0.

This is a false statement because there is no maximum point

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