Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
The graph could represent the data shown in the table is <span>the graph goes perfectly diagonal starting from zero, up to the other corner of the graph</span>
Answer:
The mean and the standard deviation of the number of students with laptops are 1.11 and 0.836 respectively.
Step-by-step explanation:
Let <em>X</em> = number of students who have laptops.
The probability of a student having a laptop is, P (X) = <em>p</em> = 0.37.
A random sample of <em>n</em> = 30 students is selected.
The event of a student having a laptop is independent of the other students.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.
The mean and standard deviation of a binomial random variable <em>X</em> are:

Compute the mean of the random variable <em>X</em> as follows:

The mean of the random variable <em>X</em> is 1.11.
Compute the standard deviation of the random variable <em>X</em> as follows:

The standard deviation of the random variable <em>X</em> is 0.836.
Answer:
1. 10 mg
2. an exponential
3. decreases only
Step-by-step explanation:
its correct
Answer:

Step-by-step explanation:
<u>Geometric Sequences</u>
There are two basic types of sequences: arithmetic and geometric. The arithmetic sequences can be recognized because each term is found as the previous term plus a fixed number called the common difference.
In the geometric sequences, each term is found by multiplying (or dividing) the previous term by a fixed number, called the common ratio.
We are given the sequence:
112, -28, 7, ...
It's easy to find out this is a geometric sequence because the signs of the terms are alternating. If it was an arithmetic sequence, the third term should be negative like the second term.
Let's find the common ratio by dividing each term by the previous term:

Testing with the third term:

Now we're sure it's a geometric sequence with r=-1/4, we use the general equation for the nth term:

