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brilliants [131]

2 years ago

9

Joshep drove his grandmother’s house which is 252 miles away in 3.5 hours. If he continues at the same speed,approximately how l

ong will It take him to drive 315 miles?

1 answer:

Amanda [17]2 years ago

5 0

Answer:

72 hours

Step-by-step explanation:

You might be interested in

Please help!!<br><br> Find the value of x

nevsk [136]

Answer:

x = 62 degrees

Step-by-step explanation:

Since this is a right angle, we can use trig functions

cos theta = adjacent/ hypotenuse

cos x = 3.7/7.9

Taking the inverse cos of each side

cos ^ -1 ( cos x) = cos ^-1 (3.7/7.9)

x =62.07246803

To the nearest degree

x = 62 degrees

5 0

2 years ago

Assume the readings on thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. Find the pro

lisov135 [29]

Answer:

0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.

The sketch is drawn at the end.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 0°C and a standard deviation of 1.00°C.

This means that $\mu = 0, \sigma = 1$

Find the probability that a randomly selected thermometer reads between −2.23 and −1.69

This is the p-value of Z when X = -1.69 subtracted by the p-value of Z when X = -2.23.

X = -1.69

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{-1.69 - 0}{1}$

$Z = -1.69$

$Z = -1.69$ has a p-value of 0.0455

X = -2.23

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{-2.23 - 0}{1}$

$Z = -2.23$

$Z = -2.23$ has a p-value of 0.0129

0.0455 - 0.0129 = 0.0326

0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.

Sketch:

3 0

2 years ago

HELP ASAP IM STUPIDDDDD <br><br><br> Write the power as a product of factors: 9^7

algol [13]

Answer:

9x9x9x9x9x9x9, theres your answer

7 0

2 years ago

Read 2 more answers

What does 10^2-2(10+5) equal

Nookie1986 [14]

Answer:70

Step-by-step explanation:

10^2 =100

-2(10+15) = -2(15)=-30

100-30= 70

4 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