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3 years ago

The graph below could be the graph of which exponential function

Mathematics

2 answers:

Irina-Kira [14]3 years ago

4 0

The correct answer is C. F(x)=2 * (0.7)^x

Anna71 [15]3 years ago

3 0

Answer:

$y=2.(0.5)^{x}$

Step-by-step explanation:

Although the options of this question have not been given, but still we can find the function from its graph.

Let the function is,

$y=a(b)^{x}$

This graph is passing through two points (0, 2) and (1, 1).

By substituting these points in the function we can find the values of a and b,

For (0, 2),

$2=a(b)^{0}$

a = 2

For (1, 1),

$1=2.(b)^{1}$

$b=\frac{1}{2}$

Therefore, given graph represents an exponential function,

$y=2.(\frac{1}{2})^{x}$

Or, $y=2.(0.5)^{x}$

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Import customs officials sometimes randomly select crates of cargo for close, but time-consuming, inspection. Suppose there are

Ostrovityanka [42]

The simple random sample is Cherryport, Dallhoise, Foxwood, and Sapphire.

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the list of random digits from left to right, starting at the beginning of the list.

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The simple random sample is Cherryport, Dallhoise, Foxwood, and Sapphire.

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2 years ago

Find the sum of the first 18 terms of the arithmetic sequence whose nth term is an = 3n - 1.

vodka [1.7K]

Answer:

495

Step-by-step explanation:

In order to find the sum of the first 18 terms you have to find the 18th term and the first term using the equation given.

a1=3(1)-1 a1=2

a18=3(18)-1 a18= 53

Then plug in 53 for an, 18 for n, and 2 in for a1 in the sum equation: Sn=n/2(a1+an)

Sn=18/2(2+53) Solve for sn= 495

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3 years ago

Suppose that W1, W2, and W3 are independent uniform random variables with the following distributions: Wi ~ Uni(0,10*i). What is

nadya68 [22]

I'll leave the computation via R to you. The $W_i$ are distributed uniformly on the intervals $[0,10i]$ , so that

$f_{W_i}(w)=\begin{cases}\dfrac1{10i}&\text{for }0\le w\le10i\\\\0&\text{otherwise}\end{cases}$

each with mean/expectation

$E[W_i]=\displaystyle\int_{-\infty}^\infty wf_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac w{10i}\,\mathrm dw=5i$

and variance

$\mathrm{Var}[W_i]=E[(W_i-E[W_i])^2]=E[{W_i}^2]-E[W_i]^2$

We have

$E[{W_i}^2]=\displaystyle\int_{-\infty}^\infty w^2f_{W_i}(w)\,\mathrm dw=\int_0^{10i}\frac{w^2}{10i}\,\mathrm dw=\frac{100i^2}3$

so that

$\mathrm{Var}[W_i]=\dfrac{25i^2}3$

Now,

$E[W_1+W_2+W_3]=E[W_1]+E[W_2]+E[W_3]=5+10+15=30$

and

$\mathrm{Var}[W_1+W_2+W_3]=E\left[\big((W_1+W_2+W_3)-E[W_1+W_2+W_3]\big)^2\right]$

$\mathrm{Var}[W_1+W_2+W_3]=E[(W_1+W_2+W_3)^2]-E[W_1+W_2+W_3]^2$

We have

$(W_1+W_2+W_3)^2={W_1}^2+{W_2}^2+{W_3}^2+2(W_1W_2+W_1W_3+W_2W_3)$

$E[(W_1+W_2+W_3)^2]$

$=E[{W_1}^2]+E[{W_2}^2]+E[{W_3}^2]+2(E[W_1]E[W_2]+E[W_1]E[W_3]+E[W_2]E[W_3])$

because $W_i$ and $W_j$ are independent when $i\neq j$ , and so

$E[(W_1+W_2+W_3)^2]=\dfrac{100}3+\dfrac{400}3+300+2(50+75+150)=\dfrac{3050}3$

giving a variance of

$\mathrm{Var}[W_1+W_2+W_3]=\dfrac{3050}3-30^2=\dfrac{350}3$

and so the standard deviation is $\sqrt{\dfrac{350}3}\approx\boxed{116.67}$

# # #

A faster way, assuming you know the variance of a linear combination of independent random variables, is to compute

$\mathrm{Var}[W_1+W_2+W_3]$

$=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]+2(\mathrm{Cov}[W_1,W_2]+\mathrm{Cov}[W_1,W_3]+\mathrm{Cov}[W_2,W_3])$

and since the $W_i$ are independent, each covariance is 0. Then

$\mathrm{Var}[W_1+W_2+W_3]=\mathrm{Var}[W_1]+\mathrm{Var}[W_2]+\mathrm{Var}[W_3]$

$\mathrm{Var}[W_1+W_2+W_3]=\dfrac{25}3+\dfrac{100}3+75=\dfrac{350}3$

and take the square root to get the standard deviation.

8 0

3 years ago

The ratio of girls to boys in art class is 4:5 if there are 32 girls how many boys are there?

dem82 [27]

Answer:

there are 40 boys

Step-by-step explanation:

32/4=8 5*8=40

8 0

3 years ago

Please answer this correctly I have to finish the sums today as soon as possible

padilas [110]

Answer:

All of them

Step-by-step explanation:

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3 0

3 years ago