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3 years ago

Pleaseee helpp ill give brainliest

Mathematics

2 answers:

vitfil [10]3 years ago

8 0

NOPE NOT AT ALL the line never goes through 0 on the x axis

satela [25.4K]3 years ago

6 0

No, (0, -2) will never appear on the line since it passes thru it in the photo

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Which of the following can be classified as Discrete and Quantitative data. The number of students in your school. The different

WITCHER [35]

Answer:

a) The number of students in your school.

Step-by-step explanation:

Quantitative and Qualitative:

The data that can be expressed with the help of numerical are know as quantitative variable.
Qualitative variable is the non parametric variable and numerical does not describe the data

Discrete and Continuous data:

Discrete data are expressed in whole number and cannot take all the values within an interval.
Continuous variable can be expressed in decimals and can take any value within an interval.

a) The number of students in your school.

Since whole numbers are used to express number of children it is a discrete and continuous data.

b) The different colors of the eyes of your classmates.

These are qualitative data and numerical are not used to express them.

c) The height of all the people in your neighborhood.

These are continuous data as height is measured and can be expressed in decimals.

d) The acceleration of your car as you drive to school.

These are continuous data as acceleration is measured and can be expressed in decimals.

3 0

3 years ago

The quotient of two rational numbers is positive. What can you conclude about the signs of the dividend and the divisor? That’s

satela [25.4K]

Answer:

The divisor and dividend have the same signs.

Step-by-step explanation:

Let's look at all of the possible outcomes of dividing with different signs.

Positive / positive = positive

Positive / negative = negative

Negative / positive = negative

Negative / negative = positive

We can see that whenever the signs are the same, the quotient is positive.

6 0

4 years ago

What is the slope of graph Pls Help

Zielflug [23.3K]

Answer:

The slope is 1/2

i believe

Step-by-step explanation:

Use the rise/run method

7 0

3 years ago

According to a poll, 80% of California adults (403 out of 506 surveyed) feel that education is one of the top issues facing Cali

fiasKO [112]

f(h(x))= 2x -21

Step-by-step explanation:

f(x)= x^3 - 6

h(x)=\sqrt[3]{2x-15}

WE need to find f(h(x)), use composition of functions

Plug in h(x)

f(h(x))=f(\sqrt[3]{2x-15})

Now we plug in f(x) in f(x)

f(h(x))=f(\sqrt[3]{2x-15})=(\sqrt[3]{2x-15})^3 - 6

cube and cube root will get cancelled

f(h(x))= 2x-15 -6= 2 x-21

6 0

3 years ago

The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(

Umnica [9.8K]

Answer:

(a) $A_1$ and $A_2$ are indeed mutually-exclusive.

(b) $\displaystyle P(A_1\; \cap \; B) = \frac{1}{20}$ , whereas $\displaystyle P(A_2\; \cap \; B) = \frac{1}{25}$ .

(d) $\displaystyle P(A_1 \; |\; B) \approx \frac{5}{9}$ , whereas $P(A_1 \; |\; B) = \displaystyle \frac{4}{9}$

Step-by-step explanation:

$P(A_1 \; \cap \; A_2) = 0$ means that it is impossible for events $A_1$ and $A_2$ to happen at the same time. Therefore, event $A_1$ and $A_2$ are mutually-exclusive.

By the definition of conditional probability:

$\displaystyle P(B \; | \; A_1) = \frac{P(B \; \cap \; A_1)}{P(B)} = \frac{P(A_1 \; \cap \; B)}{P(B)}$ .

Rearrange to obtain:

$\displaystyle P(A_1 \; \cap \; B) = P(B \; |\; A_1) \cdot P(A_1) = 0.25 \times 0.20 = \frac{1}{20}$ .

Similarly:

$\displaystyle P(A_2 \; \cap \; B) = P(B \; |\; A_2) \cdot P(A_2) = 0.80 \times 0.05 = \frac{1}{25}$ .

Note that:

$\begin{aligned}P(A_1 \; \cup \; A_2) &= P(A_1) + P(A_2) - P(A_1 \; \cap \; A_2) = 0.20 + 0.80 = 1\end{aligned}$ .

In other words, $A_1$ and $A_2$ are collectively-exhaustive. Since $A_1$ and $A_2$ are collectively-exhaustive and mutually-exclusive at the same time:

$\displaystyle P(B) = P(B \; \cap \; A_1) + P(B \; \cap \; A_2) = \frac{1}{20} + \frac{1}{25} = \frac{9}{100}$ .

By Bayes' Theorem:

$\begin{aligned} P(A_1 \; |\; B) &= \frac{P(B \; | \; A_1) \cdot P(A_1)}{P(B)} \\ &= \frac{0.25 \times 0.20}{9/100} = \frac{0.05 \times 100}{9} = \frac{5}{9}\end{aligned}$ .

Similarly:

$\begin{aligned} P(A_2 \; |\; B) &= \frac{P(B \; | \; A_2) \cdot P(A_2)}{P(B)} \\ &= \frac{0.05 \times 0.80}{9/100} = \frac{0.04 \times 100}{9} = \frac{4}{9}\end{aligned}$ .

6 0

3 years ago