All categories

2 years ago

Given a temperature of 95°C, find the equivalent in °F. 95°C = °F

Mathematics

2 answers:

r-ruslan [8.4K]2 years ago

4 0

Answer: 203°F

Step-by-step explanation: 9/5•C°+32°=F°

9/5•95°C+32°= 171°+32°=203°F

jenyasd209 [6]2 years ago

4 0

Following are the conversion of the temperature °C to °F:

Given:

95°C

To find:

°F =?

Solution:

Using formula:

$^{\circ}F = \frac{9}{5} ( ^{\circ} C) + 32$

Putting the given question value into the above formula:

$\to ^{\circ}F = \frac{9}{5} ( 95^{\circ} C) + 32\\\\\to ^{\circ}F = \frac{9}{5} \times 95^{\circ} C + 32\\\\\to ^{\circ}F = 9 \times 19^{\circ} C + 32\\\\\to ^{\circ}F = 171^{\circ} C + 32\\\\\to ^{\circ}F = 203^{\circ} C \\\\$

Therefore the final answer is " $203 ^{\circ} F$ ".

Learn more:

brainly.com/question/25428889

You might be interested in

there are 341 pine trees and spruce trees in woodlot. 122 trees were pine. what fraction of the trees were spruce?

Olenka [21]

Answer:

219 over 1000

Step-by-step explanation:

3 0

2 years ago

An artist makes a cube-shaped sculpture.

ratelena [41]

All the sides of a cube are always the same.
Area = 12 in x 12 in = 144 in^2
Volume = 12 in x 12 in x 12 in = 1728 in^3

Hence, you can confirm all the side of this cube is 12 in.

5 0

2 years ago

Cindy is twice as old as Peter. In 6 years, their ages will total 60. How old is Cindy now?

ehidna [41]

Answer:

cindny=32

peter=16

Step-by-step explanation: i think hope it helps

cindny=32

peter=16

7 0

2 years ago

If X and Y are independent continuous positive random

Leni [432]

a) $Z=\frac XY$ has CDF

$F_Z(z)=P(Z\le z)=P(X\le Yz)=\displaystyle\int_{\mathrm{supp}(Y)}P(X\le yz\mid Y=y)P(Y=y)\,\mathrm dy$

$F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}P(X\le yz)P(Y=y)\,\mathrm dy$

where the last equality follows from independence of $X,Y$ . In terms of the distribution and density functions of $X,Y$ , this is

$F_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy$

Then the density is obtained by differentiating with respect to $z$ ,

$f_Z(z)=\displaystyle\frac{\mathrm d}{\mathrm dz}\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy=\int_{\mathrm{supp}(Y)}yf_X(yz)f_Y(y)\,\mathrm dy$

b) $Z=XY$ can be computed in the same way; it has CDF

$F_Z(z)=P\left(X\le\dfrac zY\right)=\displaystyle\int_{\mathrm{supp}(Y)}P\left(X\le\frac zy\right)P(Y=y)\,\mathrm dy$

$F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}F_X\left(\frac zy\right)f_Y(y)\,\mathrm dy$

Differentiating gives the associated PDF,

$f_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}\frac1yf_X\left(\frac zy\right)f_Y(y)\,\mathrm dy$

Assuming $X\sim\mathrm{Exp}(\lambda_x)$ and $Y\sim\mathrm{Exp}(\lambda_y)$ , we have

$f_{Z=\frac XY}(z)=\displaystyle\int_0^\infty y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy$

$\implies f_{Z=\frac XY}(z)=\begin{cases}\frac{\lambda_x\lambda_y}{(\lambda_xz+\lambda_y)^2}&\text{for }z\ge0\\0&\text{otherwise}\end{cases}$

and

$f_{Z=XY}(z)=\displaystyle\int_0^\infty\frac1y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy$

$\implies f_{Z=XY}(z)=\lambda_x\lambda_y\displaystyle\int_0^\infty\frac{e^{-\lambda_x\frac zy-\lambda_yy}}y\,\mathrm dy$

I wouldn't worry about evaluating this integral any further unless you know about the Bessel functions.

6 0

2 years ago

Which equation has the same y-intercept as y=4x-3?

Masja [62]

The first one is D. Y=-3+8x
The second one is G. -3/2x=y+2

3 0

2 years ago

Read 2 more answers