Ask question

Ask question

All categories

3 years ago

6

The temperature dropped 5 °F every hour for 5 hours. What was the total number of degrees the temperature changed in the 5 hours

?

2 answers:

olasank [31]3 years ago

8 0

Answer:

it changed by 25 degrees

Step-by-step explanation:

for every hour it goes down by 5 and you have five hours so its

5+5+5+5+5

so basically its 5×5 and that equals 25

solong [7]3 years ago

4 0

Answer:

25

Step-by-step explanation:

you just multiple 5 times 5

You might be interested in

Can somebody please answers this thank you , just a, b, and c .

Shkiper50 [21]

Answer:

A) - 8 - 13 +4

B)+4 +7 - 2

C)-7 +3 - 20

5 0

3 years ago

SI TENGO 5 L DE AGUA A 5,7O Y 500ML DE ACEITE A 1,25 ¿CUANTO VALE EL LITRO DE ACEITE Y EL LITRO DE AGUA? REDONDEA A LAS CENTESIM

LuckyWell [14K]

Answer:

use a translation

Step-by-step explanation:

sorry

7 0

3 years ago

Use “decomposing” to answer the questions below

Degger [83]

A) 36
B)6
If you need help with More math let me know I’m a straight A student in math

3 0

3 years ago

(a) A lamp has two bulbs of a type with an average lifetime of 1600 hours. Assuming that we can model the probability of failure

lara [203]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The probability is $P_T= 0.4560$

b

The probability is $P_F= 0.0013$

Step-by-step explanation:

From the question we are told that

The mean for the exponential density function of bulbs failure is $\mu = 1600 \ hours$

Generally the cumulative distribution for exponential distribution is mathematically represented as

$1 - e^{- \lambda x}$

The objective is to obtain the p=probability of the bulbs failure within 1800 hours

So for the first bulb the probability will be

$P_1(x < 1800)$

And for the second bulb the probability will be

$P_2 (x< 1800)$

So from our probability that we are to determine the area to the left of 1800 on the distribution curve

Now the rate parameter $\lambda$ is mathematically represented as

$\lambda$ $= \frac{1}{\mu}$

$\lambda = \frac{1}{1600}$

The probability of the first bulb failing with 1800 hours is mathematically evaluated as

$P_1(x < 1800) = 1 - e^{\frac{1}{1600} * 1800 }$

$= 0.6753$

Now the probability of both bulbs failing would be

$P_T=P_1(x < 1800) * P_2(x < 1800)$

$= 0.6375 * 06375$

$P_T= 0.4560$

Let assume that one bulb failed at time $T_a$ and the second bulb failed at time $T_b$ then

$T_a + T_b = 1800\ hours$

The mathematical expression to obtain the probability that the first bulb failed within between zero and $T_a$ and the second bulb failed between $T_a \ and \ 1800$ is represented as

$P_F=\int_{0}^{1800}\int_{0}^{1800-x} \f{\lambda }^{2}e^{-\lambda x}* e^{-\lambda y}dx dy$

$=\int_{0}^{1800} {\lambda }e^{-\lambda x}\int_{0}^{1800-x} {\lambda } e^{-\lambda y}dx dy$

$=\int_{0}^{1800} {\frac{1}{1600} }e^{-\lambda x}\int_{0}^{1800-x} \frac{1}{1600 } e^{-\lambda y}dx dy$

$=\int_{0}^{1800} {\frac{1}{1600} }e^{-\lambda x}[e^{- \lambda y}]\left {1800-x} \atop {0}} \right. dx$

$=\int_{0}^{1800} {\frac{1}{1600} }e^{-\frac{x}{1600} }[e^{- \frac{1800 -x}{1600} }-1] dx$

$=[ {\frac{1}{1600} }e^{-\frac{1800}{1600} }-\frac{1}{1600}[e^{- \frac{x}{1600} }] \left {1800} \atop {0}} \right.$

$=[ {\frac{1}{1600} }e^{-\frac{1800}{1600} }-\frac{1}{1600}[e^{- \frac{1800}{1600} }] -[[ {\frac{1}{1600} }e^{-\frac{1800}{1600} }-\frac{1}{1600}[e^{-0}]$

$=[\frac{1}{1600} e^{-\frac{1800}{1600} } - \frac{1}{1600} e^{-0} ]$

$=0.001925 -0.000625$

$P_F= 0.0013$

4 0

3 years ago

Read 2 more answers

Choose the function whose graph is given by:

Sauron [17]

The correct answer is b or y=sin(x-2) I think. Lol dumb trigonometry

3 0

3 years ago