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3 years ago

HELP ME PLEASEEEEEEEEEEEEEEEEE

Mathematics

2 answers:

Aleks04 [339]3 years ago

8 0

Answer:

Well if i am completely correct that means the same as median so im gonna go with 8.Good luck tho

spin [16.1K]3 years ago

5 0

Second quartile is another name for median, so the answer is 8. I think. I'm second guessing myself now..

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Alexis bought 1_2 pound of grapes. How many ounces of grapes did she buy?

Olin [163]

Answer:

8 oz.

Step-by-step explanation:

One pound is 16oz and ½ of that is 8oz

6 0

3 years ago

Please assist me with this problem

Nimfa-mama [501]

Answer:

d. about 50 times larger

Step-by-step explanation:

The given expressions for magnitude (M) can be solved for the intensity (I). Then the ratio of intensities is ...

$\dfrac{I_2}{I_1}=\dfrac{I_0\cdot 10^{M_2}}{I_0\cdot 10^{M_1}}=10^{M_2-M_1}=10^{4.2-2.5}\\\\=10^{1.7}\approx 50$

The larger earthquake had about 50 times the intensity of the smaller one.

8 0

3 years ago

Which of the following answers is 5/25 simplified?<br> 1/5<br> 5/5<br> 2/5<br> 1/25

scZoUnD [109]

Answer:

1/5

Step-by-step explanation:

5/25

Divide top and bottom by 5, you get

5/25 ÷ 5/5 =1 /5

4 0

4 years ago

Read 2 more answers

The equation cosx+sinx= √3+1/2 is an identity. true or false

Mama L [17]

False - Correct on apex!!!

6 0

3 years ago

Read 2 more answers

A lamp has two bulbs, each of a type with average lifetime 1400 hours. Assuming that we can model the probability of failure of

liq [111]

Answer:

The probability of failure of both the bulbs is 0.4323.

Step-by-step explanation:

For an exponential distribution the distribution is given by

$f(x,\lambda )=\int_{0}^{x }\lambda e^{-\lambda x}dx$

The value of λ is related to the mean μ as λ=1/μ,

Let us denote the 2 bulbs by X and Y thus the probability distribution of the 2 bulbs is as under

$P(X)=\int_{0}^{x }\lambda _{X}e^{-\lambda _{X}x}dx$

Similarly for the bulb Y the distribution function is given by

$P(Y)=\int_{0}^{y }\lambda _{Y}e^{-\lambda _{Y}y}dy$

Thus the probability for both the bulbs to fail within 1500 hours is

$P(E)=\int_{0}^{1500}\int_{0}^{1500}\frac{1}{1400}e^{\frac{-x}{1400}}\cdot \frac{1}{1400}e^{\frac{-y}{1400}}dxdy\\\\P(E)=\frac{1}{1400^2}(\int_{0}^{1500}\int_{0}^{1500}e^{\frac{-x}{1400}}\cdot e^{\frac{-y}{1400}}dxdy)\\\\P(E)=\frac{1}{1400^2}(\int_{0}^{1500}e^{\frac{-x}{1400}}dx)\cdot (\int_{0}^{1500}e^{\frac{-y}{1400}}dy)\\\\P(E)=\frac{1}{1400^{2}}\times 920.473\times 920.473\\\\\therefore P(E)=0.4323$

6 0

3 years ago