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3 years ago

Two mechanics worked on a car. The first mechanic worked for 5

Mathematics

1 answer:

ArbitrLikvidat [17]3 years ago

8 0

Answer:

$50 and $115

Step-by-step explanation:

We know that,

the sum of the two rates = $165 per hour

Let x and (165 - x) represent the $/hour of the 15hr and 5hr mechanics respectively.

Then combining their charges together to get:

$(x*15)hour+(165-x)*5 hour=1325$

Solving for x to get:

$15x-5x+825=1325$

$10x=500$

$x=50$

So the rate charged per hour by one mechanic is $50 and for the other mechanic = $(165-50)=$ $115.

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(X^2+y^2+x)dx+xydy=0 Solve for general solution

aksik [14]

Check if the equation is exact, which happens for ODEs of the form

$M(x,y)\,\mathrm dx+N(x,y)\,\mathrm dy=0$

if $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ .

We have

$M(x,y)=x^2+y^2+x\implies\dfrac{\partial M}{\partial y}=2y$

$N(x,y)=xy\implies\dfrac{\partial N}{\partial x}=y$

so the ODE is not quite exact, but we can find an integrating factor $\mu(x,y)$ so that

$\mu(x,y)M(x,y)\,\mathrm dx+\mu(x,y)N(x,y)\,\mathrm dy=0$

is exact, which would require

$\dfrac{\partial(\mu M)}{\partial y}=\dfrac{\partial(\mu N)}{\partial x}\implies \dfrac{\partial\mu}{\partial y}M+\mu\dfrac{\partial M}{\partial y}=\dfrac{\partial\mu}{\partial x}N+\mu\dfrac{\partial N}{\partial x}$

$\implies\mu\left(\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}\right)=M\dfrac{\partial\mu}{\partial y}-N\dfrac{\partial\mu}{\partial x}$

Notice that

$\dfrac{\partial N}{\partial x}-\dfrac{\partial M}{\partial y}=y-2y=-y$

is independent of x, and dividing this by $N(x,y)=xy$ gives an expression independent of y. If we assume $\mu=\mu(x)$ is a function of x alone, then $\frac{\partial\mu}{\partial y}=0$ , and the partial differential equation above gives

$-\mu y=-xy\dfrac{\mathrm d\mu}{\mathrm dx}$

which is separable and we can solve for $\mu$ easily.

$-\mu=-x\dfrac{\mathrm d\mu}{\mathrm dx}$

$\dfrac{\mathrm d\mu}\mu=\dfrac{\mathrm dx}x$

$\ln|\mu|=\ln|x|$

$\implies \mu=x$

So, multiply the original ODE by x on both sides:

$(x^3+xy^2+x^2)\,\mathrm dx+x^2y\,\mathrm dy=0$

Now

$\dfrac{\partial(x^3+xy^2+x^2)}{\partial y}=2xy$

$\dfrac{\partial(x^2y)}{\partial x}=2xy$

so the modified ODE is exact.

Now we look for a solution of the form $F(x,y)=C$ , with differential

$\mathrm dF=\dfrac{\partial F}{\partial x}\,\mathrm dx+\dfrac{\partial F}{\partial y}\,\mathrm dy=0$

The solution F satisfies

$\dfrac{\partial F}{\partial x}=x^3+xy^2+x^2$

$\dfrac{\partial F}{\partial y}=x^2y$

Integrating both sides of the first equation with respect to x gives

$F(x,y)=\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+f(y)$

Differentiating both sides with respect to y gives

$\dfrac{\partial F}{\partial y}=x^2y+\dfrac{\mathrm df}{\mathrm dy}=x^2y$

$\implies\dfrac{\mathrm df}{\mathrm dy}=0\implies f(y)=C$

So the solution to the ODE is

$F(x,y)=C\iff \dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3+C=C$

$\implies\boxed{\dfrac{x^4}4+\dfrac{x^2y^2}2+\dfrac{x^3}3=C}$

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3 years ago

4r+3=-18 please help me

lorasvet [3.4K]

Subtract three from either side. That will give you 4r=-21. Divide -21 by 4 and r will equal -5.25.

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3 years ago

write the number that can be shown with this many hundreds, tens, and ones, 1 hundreds, 2 tens, 18 ones

NARA [144]

It is 138 who won the know that

5 0

3 years ago

Suppose a geyser has a mean time between eruptions of 72 minutes. Let the interval of time between the eruptions be normally dis

nikitadnepr [17]

Answer:

(a) The probability that a randomly selected time interval between eruptions is longer than 82 minutes is 0.3336.

(b) The probability that a random sample of 13-time intervals between eruptions has a mean longer than 82 minutes is 0.0582.

(c) The probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 minutes is 0.0055.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) The population mean must be more than 72, since the probability is so low.

Step-by-step explanation:

We are given that a geyser has a mean time between eruptions of 72 minutes.

Also, the interval of time between the eruptions be normally distributed with a standard deviation of 23 minutes.

(a) Let X = the interval of time between the eruptions

So, X ~ N( $\mu=72, \sigma^{2} =23^{2}$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time = 72 minutes

$\sigma$ = standard deviation = 23 minutes

Now, the probability that a randomly selected time interval between eruptions is longer than 82 minutes is given by = P(X > 82 min)

P(X > 82 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{82-72}{23}$ ) = P(Z > 0.43) = 1 - P(Z $\leq$ 0.43)

= 1 - 0.6664 = 0.3336

The above probability is calculated by looking at the value of x = 0.43 in the z table which has an area of 0.6664.

(b) Let $\bar X$ = sample mean time between the eruptions

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time = 72 minutes

$\sigma$ = standard deviation = 23 minutes

n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between eruptions has a mean longer than 82 minutes is given by = P( $\bar X$ > 82 min)

P( $\bar X$ > 82 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{82-72}{\frac{23}{\sqrt{13} } }$ ) = P(Z > 1.57) = 1 - P(Z $\leq$ 1.57)

= 1 - 0.9418 = 0.0582

The above probability is calculated by looking at the value of x = 1.57 in the z table which has an area of 0.9418.

(c) Let $\bar X$ = sample mean time between the eruptions

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time = 72 minutes

$\sigma$ = standard deviation = 23 minutes

n = sample of time intervals = 34

Now, the probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 minutes is given by = P( $\bar X$ > 82 min)

P( $\bar X$ > 82 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{82-72}{\frac{23}{\sqrt{34} } }$ ) = P(Z > 2.54) = 1 - P(Z $\leq$ 2.54)

= 1 - 0.9945 = 0.0055

The above probability is calculated by looking at the value of x = 2.54 in the z table which has an area of 0.9945.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) If a random sample of 34-time intervals between eruptions has a mean longer than 82 minutes, then we conclude that the population mean must be more than 72, since the probability is so low.

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3 years ago

How to solve 10 (g+5)=2(g+9)

Ghella [55]

first multiply the outside numbers by both the numbers in the ()

10xg=10g 10x5=50 2xg=2g 2x9=18

then combine like terms

10g+50=2g+18

8g=32

then divide

g=4

4 0

3 years ago