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3 years ago

What is the mode? 0-0-2-1-3-2-1-0

Mathematics

2 answers:

exis [7]3 years ago

4 0

The mode is 0 since it is the most frequent.

Tanya [424]3 years ago

4 0

The mode is 0 shjsjs

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3 years ago

In a certain Algebra 2 class of 29 students, 7 of them play basketball and 24 of them play baseball. There are 3 students who pl

antiseptic1488 [7]

Answer:

$\frac{5}{29}$

Step-by-step explanation:

Let $n(A)$ represent students playing basketball, $n(B)$ represent students playing baseball.

Then, $n(A)=7$ , $n(B)=24$

Let $n(S)$ be the total number of students. So, $n(S)=29$ .

Now,

$P(A)=\frac{n(A)}{n(S)}=\frac{7}{29}$

$P(B)=\frac{n(B)}{n(S)}=\frac{24}{29}$

3 students play neither of the sport. So, students playing either of the two sports is given as:

$n(A\cup B)=n(S)-3\\n(A\cup B)=29-3=26$

∴ $P(A\cup B)=\frac{n(A\cup B)}{n(S)}=\frac{26}{29}$

From the probability addition theorem,

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Where, $P(A\cap B)$ is the probability that a student chosen randomly from the class plays both basketball and baseball.

Plug in all the values and solve for $P(A\cap B)$ . This gives,

$\frac{26}{29}=\frac{7}{29}+\frac{24}{29}+P(A\cap B)\\\\\frac{26}{29}=\frac{7+24}{29}+P(A\cap B)\\\\\frac{26}{29}=\frac{31}{29}+P(A\cap B)\\\\P(A\cap B=\frac{31}{29}-\frac{26}{29}\\\\P(A\cap B=\frac{31-26}{29}=\frac{5}{29}$

Therefore, the probability that a student chosen randomly from the class plays both basketball and baseball is $\frac{5}{29}$

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3 years ago

A circle is described by the equation (x−1/2 )^2 +(y−3/2 )^2 =4/9 . What are the cordinates for the center of the circle and the

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3 years ago

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-13.4 is the answer. And this is simplified form

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3 years ago

Read 2 more answers

The parent function f(x) = x3 is translated to form g(x), as shown on the graph. The translated function can be written in the f

OLEGan [10]

Answer: First option

Step-by-step explanation:

By definition, the horizontal shift depends on the value of h and the vertical shift depends on the value of k.

$f(x)=(x-h)^3$ indicates that the function if shifted to the right h units.

$f(x)=(x+h)^3$ indicates that the function if shifted to the left h units.

$f(x)=f(x)-k$ indicates that the function if shifted down k units.

$f(x)=f(x)+k$ indicates that the function if shifted up k units.

Then:

If h is positive, the graph will shift to the right.

If k is negative, the graph will shift down.

As you can see in the graph, the function is shifted 4 units to the right and 2 units down. Therefore g(x) has the form:

$g(x)=(x-4)^3-2$

Where:

$h=4\\k=-2$

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3 years ago