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3 years ago

Which graphs show continuous data? Select each correct answer.

Mathematics

2 answers:

love history [14]3 years ago

5 0

C

Due to the fact that it is starting from 0 and moving in a steady pattern

Shalnov [3]3 years ago

4 0

Answer:

B and D

Step-by-step explanation:

I just took the test.

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Tema [17]

Maybe B (not sure).................

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3 years ago

travel earns $25 a day working afther school at the supermarket. how much money will he earn in d days

hichkok12 [17]

Answer:

Hi!

Your answer is:

25d

Since we aren't aware of how many days he is working, we substitute a variable (d) in for it!

I hope this helps!

4 0

4 years ago

List the interior angles shown in the figure

Oksi-84 [34.3K]

Answer:

The angles are 5,2, and 3.

Step-by-step explanation:

Well, they're inside the triangle.

3 0

3 years ago

Read 2 more answers

Which of the following are incorrect expressions for slope?

vfiekz [6]

Answer:

Option B and C are correct.

$\frac{x_2-x_1}{y_2-y_1}$

$\frac{run}{rise}$ are the expression incorrect for slope

Step-by-step explanation:

Slope is defined as the change in the dependent variable relative to the change in the dependent variable

or the ratio of the horizontal changes to vertical changes between any two points on the graph of the line.

The vertical changes between any two points is rise

The horizontal changes between any two points is run.

Formula for slope is given by:

For any two points $(x_1, y_1)$ and $(x_2, y_2)$

then slope is:

$\text{Slope} =\frac{rise}{run}= \frac{y_2-y_1}{x_2-x_1}$

or we can write this as:

Δy = $y_2-y_1$

Δx = $x_2-x_1$

⇒ $\text{Slope} = \frac{\triangle y}{\triangle x}$

Therefore, the expression which are incorrect for slope are;

$\frac{x_2-x_1}{y_2-y_1}$

$\frac{run}{rise}$

8 0

3 years ago

Read 2 more answers

41% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the

lys-0071 [83]

Answer:

a) 0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly five.

b) 0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least six.

c) 0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

Step-by-step explanation:

For each adult, there are only two possible outcomes. Either they have very little confidence in newspapers, or they do not. The answers of each adult are independent, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

41% of U.S. adults have very little confidence in newspapers.

This means that $p = 0.41$

You randomly select 10 U.S. adults.

This means that $n = 10$

(a) exactly five

This is $P(X = 5)$ . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 5) = C_{10,5}.(0.41)^{5}.(0.59)^{5} = 0.2087$

0.2087 = 20.82% probability that the number of U.S. adults who have very little confidence in newspapers is exactly five.

(b) at least six

This is:

$P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{10,6}.(0.41)^{6}.(0.59)^{4} = 0.1209$

$P(X = 7) = C_{10,7}.(0.41)^{7}.(0.59)^{3} = 0.0480$

$P(X = 8) = C_{10,8}.(0.41)^{8}.(0.59)^{2} = 0.0125$

$P(X = 9) = C_{10,9}.(0.41)^{9}.(0.59)^{1} = 0.0019$

$P(X = 10) = C_{10,10}.(0.41)^{10}.(0.59)^{0} = 0.0001$

Then

$P(X \geq 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.1209 + 0.0480 + 0.0125 + 0.0019 + 0.0001 = 0.1834$

0.1834 = 18.34% probability that the number of U.S. adults who have very little confidence in newspapers is at least six.

(c) less than four.

This is:

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.41)^{0}.(0.59)^{10} = 0.0051$

$P(X = 1) = C_{10,1}.(0.41)^{1}.(0.59)^{9} = 0.0355$

$P(X = 2) = C_{10,2}.(0.41)^{2}.(0.59)^{8} = 0.1111$

$P(X = 3) = C_{10,3}.(0.41)^{3}.(0.59)^{7} = 0.2058$

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0051 + 0.0355 + 0.1111 + 0.2058 = 0.3575$

0.3575 = 35.75% probability that the number of U.S. adults who have very little confidence in newspapers is less than four.

5 0

3 years ago