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erastova [34]
3 years ago
5

What is 8 times 10 to the third power

Mathematics
2 answers:
Liula [17]3 years ago
7 0
8 times ten to the third power is 8000
I am Lyosha [343]3 years ago
5 0
8000 is the answer of the problem
You might be interested in
3/4 of a piece of metal has a mass of 15kg.what is the mass of 2/5 of the piece of metal
olga2289 [7]
3/4=15
let,
total mass is x
x=4×15/3
=20 kg
so, total mass = 20 kg

=2×20/5
=2×4
=8 kg

Hope this helps!
8 0
3 years ago
Can the government offices be represented as a function of the ministers' names?
Marina86 [1]

Answer and explanation:

A function is given f(x)= y where x is independent variable and y is dependent variable. Y is a function of x therefore y takes up a value because x has a certain value, y is dependent on x values.

Therefore if government office y is a function of minister's name x then a government office would hold a particular value, example be named after the minister's name because it is dependent on the minister's name x

6 0
3 years ago
Show that if X is a geometric random variable with parameter p, then
Lubov Fominskaja [6]

Answer:

\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}

Step-by-step explanation:

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

P(X=x)=(1-p)^{x-1} p

Let X the random variable that measures the number os trials until the first success, we know that X follows this distribution:

X\sim Geo (1-p)

In order to find the expected value E(1/X) we need to find this sum:

E(X)=\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}

Lets consider the following series:

\sum_{k=1}^{\infty} b^{k-1}

And let's assume that this series is a power series with b a number between (0,1). If we apply integration of this series we have this:

\int_{0}^b \sum_{k=1}^{\infty} r^{k-1}=\sum_{k=1}^{\infty} \int_{0}^b r^{k-1} dt=\sum_{k=1}^{\infty} \frac{b^k}{k}   (a)

On the last step we assume that 0\leq r\leq b and \sum_{k=1}^{\infty} r^{k-1}=\frac{1}{1-r}, then the integral on the left part of equation (a) would be 1. And we have:

\int_{0}^b \frac{1}{1-r}dr=-ln(1-b)

And for the next step we have:

\sum_{k=1}^{\infty} \frac{b^{k-1}}{k}=\frac{1}{b}\sum_{k=1}^{\infty}\frac{b^k}{k}=-\frac{ln(1-b)}{b}

And with this we have the requiered proof.

And since b=1-p we have that:

\sum_{k=1}^{\infty} \frac{p(1-p)^{k-1}}{k}=-\frac{p ln p}{1-p}

4 0
3 years ago
Assume that a hypothesis test of the given claim will be conducted. Identify the type I or type II error for the test. A consume
Ivan

Answer:

Type I error: Concluding that mean mileage is less than 32 miles per hour when actually it is greater than or equal to 32 miles per gallon.

Step-by-step explanation:

We are given the following in the question:

Hypothesis:

Mean mileage for the Carter Motor​ Company's new sedan

We can design the null hypothesis and alternate hypothesis as:

H_{0}: \mu \geq 32\text{ miles per gallon}\\H_A: \mu < 32\text{ miles per gallon}

Type I error:

  • It is the false positive error.
  • It is the error of rejection a true hypothesis.

Type II error:

  • It is the false negative error.
  • It is the non rejection of a false null hypothesis.

Thus, type I error for the given hypothesis is concluding that mean mileage is less than 32 miles per hour when actually it is greater than or equal to 32 miles per gallon.

Type II error would be concluding that mean mileage is greater than or equal to 32 miles per gallon when actually it is less than 32 miles per gallon.

3 0
3 years ago
What represents the opposite of the opposite of 8 1/4
satela [25.4K]
The answer would be 8 1/4 because if you have the opposite of the opposite you get what you started with.  For example the opposite of night is day then the opposite of day is night so you end up with what you started with.
8 0
3 years ago
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