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3 years ago

Ill mark you brainlist plz help if you put something random I will report you!

Mathematics

2 answers:

Serhud [2]3 years ago

6 0

Answer:

Step-by-step explanation:

$-10, -7, -4, -1, 2, 5,$

Common difference = +3

Next term = 2 + 3 = 5

Nadya [2.5K]3 years ago

3 0

Answer:

3 is the common difference

The next term in the sequence would be 5

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V≈56.55in³

Step-by-step explanation:

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3 years ago

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Can anyone help me with the pictures below? Will mark brainliest!

Ira Lisetskai [31]

Answer:

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4 years ago

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4 years ago

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

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

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3 years ago

Preview Activity 3.5.1. A spherical balloon is being inflated at a constant rate of 20 cubic inches per second. How fast is the

ArbitrLikvidat [17]

Answer:

The radius is increasing at a rate of approximately 0.044 in/s when the diameter is 12 inches.

Because $\frac{dr}{dt}=\frac{5}{36\pi }>\frac{dr}{dt}=\frac{5}{64\pi }$ the radius is changing more rapidly when the diameter is 12 inches.

Step-by-step explanation:

Let $r$ be the radius, $d$ the diameter, and $V$ the volume of the spherical balloon.

We know $\frac{dV}{dt}=20 \:{in^3/s}$ and we want to find $\frac{dr}{dt}$

The volume of a spherical balloon is given by

$V=\frac{4}{3} \pi r^3$

Taking the derivative with respect of time of both sides gives

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$

We now substitute the values we know and we solve for $\frac{dr}{dt}$ :

$d=2r\\\\r=\frac{d}{2}$

$r=\frac{12}{2}=6$

$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\\\\\frac{dr}{dt}=\frac{\frac{dV}{dt}}{4\pi r^2} \\\\\frac{dr}{dt}=\frac{20}{4\pi(6)^2 } =\frac{5}{36\pi }\approx 0.044$

The radius is increasing at a rate of approximately 0.044 in/s when the diameter is 12 inches.

When d = 16, r = 8 and $\frac{dr}{dt}$ is:

$\frac{dr}{dt}=\frac{20}{4\pi(8)^2}=\frac{5}{64\pi }\approx 0.025$

The radius is increasing at a rate of approximately 0.025 in/s when the diameter is 16 inches.

Because $\frac{dr}{dt}=\frac{5}{36\pi }>\frac{dr}{dt}=\frac{5}{64\pi }$ the radius is changing more rapidly when the diameter is 12 inches.

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3 years ago