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Murljashka [212]
3 years ago
11

The temperature of a liquid in an experiment changes by −10.4°F from the beginning of the experiment to the first check. After m

aking some adjustments, the scientist checks the temperature again later and finds that it is 17.7°F, which is 13 what it was at the time of the first check. What was the initial temperature of the liquid?
Mathematics
1 answer:
ziro4ka [17]3 years ago
3 0

Answer:    63.5 F

Step-by-step explanation:

you are correct up there I took the test and it was correct

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Find all solutions to the equation in the interval [0, 2π).
ANEK [815]

Answer:

x = π/6, π/2, 5π/6, 3π/2

Step-by-step explanation:

cos x = sin(2x)

Use double angle formula.

cos x = 2 sin x cos x

Move everything to one side and factor.

cos x − 2 sin x cos x = 0

cos x (1 − 2 sin x) = 0

Set each factor to 0 and solve.

cos x = 0

x = π/2, 3π/2

1 − 2 sin x = 0

sin x = 1/2

x = π/6, 5π/6

The total solution is:

x = π/6, π/2, 5π/6, 3π/2

6 0
4 years ago
How many solutions does the following equation have?<br> |3x + 12| = 18 (5 points)
DedPeter [7]

Answer: 2 solutions.

The given equation breaks down into these two equations

3x+12 = 18 and 3x+12 = -18

Basically 3x+12 equal to plus 18 and minus 18. This is due to the fact that something like |x| = 7 means x = 7 or x = -7, ie those x values are exactly seven units away from zero on the number line.

If you want to solve each equation, then

3x+12 = 18

3x = 18-12

3x = 6

x = 6/3

x = 2 is one solution

and

3x+12 = -18

3x = -18-12

3x = -30

x = -30/3

x = -10 is the other solution

5 0
3 years ago
Suppose X and Y are random variables with joint density function. f(x, y) = 0.1e−(0.5x + 0.2y) if x ≥ 0, y ≥ 0 0 otherwise (a) I
Hatshy [7]

a. f_{X,Y} is a joint density function if its integral over the given support is 1:

\displaystyle\int_{-\infty}^\infty\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\frac1{10}\int_0^\infty\int_0^\infty e^{-x/2-y/5}\,\mathrm dx\,\mathrm dy

=\displaystyle\frac1{10}\left(\int_0^\infty e^{-x/2}\,\mathrm dx\right)\left(\int_0^\infty e^{-y/5}\,\mathrm dy\right)=\frac1{10}\cdot2\cdot5=1

so the answer is yes.

b. We should first find the density of the marginal distribution, f_Y(y):

f_Y(y)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx=\frac1{10}\int_0^\infty e^{-x/2-y/5}\,\mathrm dy

f_Y(y)=\begin{cases}\dfrac15e^{-y/5}&\text{for }y\ge0\\\\0&\text{otherwise}\end{cases}

Then

P(Y\ge8)=\displaystyle\int_8^\infty f_Y(y)\,\mathrm dy=e^{-8/5}

or about 0.2019.

For the other probability, we can use the joint PDF directly:

P(X\le5,Y\le8)=\displaystyle\int_0^5\int_0^8f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=1+e^{-41/10}-e^{-5/2}-e^{-8/5}

which is about 0.7326.

c. We already know the PDF for Y, so we just integrate:

E[Y]=\displaystyle\int_{-\infty}^\infty y\,f_Y(y)\,\mathrm dy=\frac15\int_0^\infty ye^{-y/5}\,\mathrm dy=\boxed5

5 0
3 years ago
a store sold a case of scented candles for$17.85 that had been marked up 110%. what was the original price
densk [106]
If 10% was added to the total (100%), then subtract 10% from 17.85.

17.85 * 0.1 = 1.785
17.85 - 1.785 = 16.065
$16.07
6 0
3 years ago
PLEASE PLEASE HELPPPP
inysia [295]

Answer:

The answer is 173.2

Step-by-step explanation:

8 0
3 years ago
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