All categories

2 years ago

Will give brainliest if correct

1 answer:

7 0

Answer: The correct answer would be choice D, as long as the t was an exponent.

It should be written as 233.10 = 102.15(1.086)^t

The t has to be an exponent because you are multiplying the investment by 1.086 repeatedly each quarter. We used 1.086 because we added on 8.6% (or 0.086) to the investment.

You might be interested in

What is 0.666667 as a fraction ?

Olenka [21]

0.666667 as a fraction is 2/3

7 0

2 years ago

Read 2 more answers

Which statement is true about a line and a point? (1 point)

Oxana [17]

A line and a point cannot be collinear.

5 0

1 year ago

TJ is thinking of a number

Likurg_2 [28]

Answer:

$\frac{1}{3}n - 5$

Step-by-step explanation:

If n is the number that TJ is thinking, then says he found 1 third of the number, you would multiply 1/3 by n. Then he says to subtract 5, which would be the -5.

8 0

1 year ago

ALGEBRA 2/ TRIG QUESTION ATTACHED Please help ❤️❤️

ss7ja [257]

Let's calculate the value of angle A and B

sin(A) =-4/5 → sin⁻¹(- 4/5) = A → A = - 53.13

cos(B) = -5/13 → cos⁻¹ (- 5/13) = B → B = 112.62

tan (A+B) = sin(A+B)/cos(A+B) with A+B = -53.13 + 112.62 = 59.49

tan (A+B) = sin(59.49)/cos(59.49) = 0.86154/0.507688 = 1.6969.

(Answer H = 56/33 = 1.6969)

7 0

2 years ago

At an ocean-side nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water t

grandymaker [24]

Answer:

(a1) The probability that temperature increase will be less than 20°C is 0.667.

(a2) The probability that temperature increase will be between 20°C and 22°C is 0.133.

(b) The probability that at any point of time the temperature increase is potentially dangerous is 0.467.

Step-by-step explanation:

Let X = temperature increase.

The random variable X follows a continuous Uniform distribution, distributed over the range [10°C, 25°C].

The probability density function of X is:

$f(X)=\left \{ {{\frac{1}{25-10}=\frac{1}{15};\ x\in [10, 25]} \atop {0;\ otherwise}} \right.$

(a1)

Compute the probability that temperature increase will be less than 20°C as follows:

$P(X$

Thus, the probability that temperature increase will be less than 20°C is 0.667.

(a2)

Compute the probability that temperature increase will be between 20°C and 22°C as follows:

$P(20$

Thus, the probability that temperature increase will be between 20°C and 22°C is 0.133.

(b)

Compute the probability that at any point of time the temperature increase is potentially dangerous as follows:

$P(X>18)=\int\limits^{25}_{18}{\frac{1}{15}}\, dx\\=\frac{1}{15}\int\limits^{25}_{18}{dx}\,\\=\frac{1}{15}[x]^{25}_{18}=\frac{1}{15}[25-18]=\frac{7}{15}\\=0.467$

Thus, the probability that at any point of time the temperature increase is potentially dangerous is 0.467.

(c)

Compute the expected value of the uniform random variable X as follows:

$E(X)=\frac{1}{2}[10+25]=\frac{35}{2}=17.5$

Thus, the expected value of the temperature increase is 17.5°C.

7 0

2 years ago