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3 years ago

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1. What is the decimal equivalent to start fraction 3 over 50 end fraction? (1 point)

1 answer:

Olegator [25]3 years ago

4 0

Answer:

Step-by-step explanation:

0.06
27/50
1/64
I am not sure but I think it is 1/36
3⁵
Idaho has the smallest population.
0.000857
1.414213562..
1.5 x 10⁸
.16
3/32
6.2
29/30
I am not sure
-1/6
4/9
2.25
.265

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Suppose the population standard deviation is σ = 5 , an SRS of n = 100 is obtained, and the confidence level is chosen to be 98%

beks73 [17]

Answer:

1.165.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.98}{2} = 0.01$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.01 = 0.99$ , so $z = 2.33$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

In this problem, we have that:

$\sigma = 5, n = 100$ . So

$M = z*\frac{\sigma}{\sqrt{n}}$

$M = 2.33*\frac{5}{\sqrt{100}} = 1.165$

So the correct answer is:

1.165.

6 0

3 years ago

Which expression represents the product of b and 18?

Anastaziya [24]

Answer:

You're answer is D.18b

6 0

1 year ago

Read 2 more answers

Numbers are (-20,10) and (40,-50)

seropon [69]

Hdbanwlsksmaakdicjcjdndisisushshsudhehehhs

6 0

2 years ago

A large corporation starts at time t = 0 to invest part of its receipts continuously at a rate of P dollars per year in a fund f

Andrews [41]

Answer:

$A = \frac{P}{r}\left( e^{rt} -1 \right)$

Step-by-step explanation:

This is <em>a separable differential equation</em>. Rearranging terms in the equation gives

$\frac{dA}{rA+P} = dt$

Integration on both sides gives

$\int \frac{dA}{rA+P} = \int dt$

where $c$ is a constant of integration.

The steps for solving the integral on the right hand side are presented below.

$\int \frac{dA}{rA+P} = \begin{vmatrix} rA+P = m \implies rdA = dm\end{vmatrix} \\\\\phantom{\int \frac{dA}{rA+P} } = \int \frac{1}{m} \frac{1}{r} \, dm \\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \int \frac{1}{m} \, dm\\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |m| + c \\\\&\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |rA+P| +c$

Therefore,

$\frac{1}{r} \ln |rA+P| = t+c$

Multiply both sides by $r.$

$\ln |rA+P| = rt+c_1, \quad c_1 := rc$

By taking exponents, we obtain

$e^{\ln |rA+P|} = e^{rt+c_1} \implies |rA+P| = e^{rt} \cdot e^{c_1} rA+P = Ce^{rt}, \quad C:= \pm e^{c_1}$

Isolate $A$ .

$rA+P = Ce^{rt} \implies rA = Ce^{rt} - P \implies A = \frac{C}{r}e^{rt} - \frac{P}{r}$

Since $A = 0$ when $t=0$ , we obtain an initial condition $A(0) = 0$ .

We can use it to find the numeric value of the constant $c$ .

Substituting $0$ for $A$ and $t$ in the equation gives

$0 = \frac{C}{r}e^{0} - \frac{P}{r} \implies \frac{P}{r} = \frac{C}{r} \implies C=P$

Therefore, the solution of the given differential equation is

$A = \frac{P}{r}e^{rt} - \frac{P}{r} = \frac{P}{r}\left( e^{rt} -1 \right)$

4 0

2 years ago

HELPPPPP PLEASEEEEE

Salsk061 [2.6K]

Answer:- 1/2

Step-by-step explanation:

4 0

2 years ago