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1 year ago

What is the common difference between the elements of the arithmetic sequence below?

2 answers:

8 0

In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. The common difference would be calculated as follows:


–36 - –31.5 = - 4.5
–27 - –22.5 = -4.5

Therefore, the common difference is -4.5.

fredd [130]1 year ago

3 0

All numbers in each number add up to 9, i dont know if you understand me,
for example:
1+8=9
2+2+5=9
2+7=9
.....

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Let X represent the amount of time until the next student will arrive in the library parking lot at the university. If we know t

Ber [7]

Answer:

The probability that it will take more than 10 minutes for the next student to arrive at the library parking lot is 0.0821.

Step-by-step explanation:

The random variable X is defined as the amount of time until the next student will arrive in the library parking lot at the university.

The random variable X follows an Exponential distribution with mean, μ = 4 minutes.

The probability density function of X is:

$f_{X}(x)=\lambda e^{\lambda x};\ x\geq 0, \lambda >0$

The parameter of the exponential distribution is:

$\lambda=\frac{1}{\mu}=\frac{1}{4}=0.25$

Compute the value of P (X > 10) as follows:

$P(X>10)=\int\limits^{\infty}_{10}{0.25e^{-0.25x}}\, dx$

$=0.25\times \int\limits^{\infty}_{10}{e^{-0.25x}}\, dx\\=0.25\times |\frac{e^{0.25x}}{-0.25}|^{\infty}_{10}\\=(e^{-0.25\times \infty})-(e^{-0.25\times 10})\\=0.0821$

Thus, the probability that it will take more than 10 minutes for the next student to arrive at the library parking lot is 0.0821.

3 0

2 years ago

The course length of the Boston Marathon is about 4.2×104 meters. The course length of the Iditarod Trail Invitational ultra-mar

GREYUIT [131]

Answer:

The ultra-marathon Idita Rod Trail is about 38 times longer than the Boston Marathon.

Step-by-step explanation:

Let

x -----> the length of the Boston Marathon in meters

y -----> the length of the Idita rod Trail Invitational ultra-marathon in meters

we have

$x=4.2*10^4\ m$

$y=1.6*10^6\ m$

we know that

To find out how many times as long is the course of the Idita rod Trail ultra-marathon as that of the Boston Marathon, divide the length of the Idita rod Trail Invitational ultra-marathon by the length of the Boston Marathon

$\frac{y}{x}$

$\frac{x}{y}=\frac{1.6*10^6}{4.2*10^4}$

Remember that

To divide two numbers in scientific notation, divide their coefficients and subtract their exponents

$\frac{1.6*10^6}{4.2*10^4}=\frac{1.6}{4.2}*(10^{6-4})=0.381*(10^{2})=38.1$

therefore

The ultra-marathon Idita Rod Trail is about 38 times longer than the Boston Marathon.

5 0

2 years ago

A tank contains 8000 L of pure water. Brine that contains 35 g of salt per liter of water is pumped into the tank at a rate of 2

OleMash [197]

Answer:

The concentration of salt in the tank approaches $35 \mathrm{g} / \mathrm{L},$

Step-by-step explanation:

Data provide in the question:

Water contained in the tank = 8000 L

Salt per litre contained in Brine = 35 g/L

Rate of pumping water into the tank = 25 L/min

Concentration of salt $\lim _{t \rightarrow \infty} C(t)=\lim _{t \rightarrow \infty} \frac{35 t}{320+t}$

Now,

Dividing both numerator and denominator by $t$ , we have

$\lim _{t \rightarrow \infty} \frac{\frac{1}{t} 35 t}{\frac{1}{t}(320+t)}=\lim _{t \rightarrow \infty} \frac{35}{\frac{320}{t}+1}=35$

Here,

The concentration of salt in the tank approaches $35 \mathrm{g} / \mathrm{L},$

3 0

2 years ago

Anita had $400 in her savings account when she went to college. Her parents will add $200 to her account each month. Miguel had

lesantik [10]

At first, Anita looks like her account would grow faster. By month 4, shes at $1,200 while Miguel is at $400. However, Anita is only getting $200 a month. While Miguel gets double of whatever amount he has in his bank account. By month 10, Anita is at $2,400 while Miguel has $25,600. That is a $23,200 difference. So Miguel's bank account will grow faster than Anita.

5 0

1 year ago

Tan 235° = 2tan20°+ tan215°

Mariulka [41]

Given : tan 235 = 2 tan 20 + tan 215

To Find : prove that

Solution:

tan 235 = 2 tan 20 + tan 215

Tan x = Tan (180 + x)

tan 235 = tan ( 180 + 55) = tan55

tan 215 = tan (180 + 35) = tan 35

=> tan 55 = 2tan 20 + tan 35

55 = 20 + 35

=> 20 = 55 - 35

taking Tan both sides

=> Tan 20 = Tan ( 55 - 35)

=> Tan 20 = (Tan55 - Tan35) /(1 + Tan55 . Tan35)

Tan35 = Cot55 = 1/tan55 => Tan55 . Tan35 =1

=> Tan 20 = (Tan 55 - Tan 35) /(1 + 1)

=> Tan 20 = (Tan 55 - Tan 35) /2

=> 2 Tan 20 = Tan 55 - Tan 35

=> 2 Tan 20 + Tan 35 = Tan 55

=> tan 55 = 2tan 20 + tan 35

=> tan 235 = 2tan 20 + tan 215

QED

Hence Proved

5 0

1 year ago