20a-18<br> 2: 42<br> 3: 17b+3<br> 4: 105

Thirty Mercedes and Audi participated in a 30 mile race. The average driving speed of the Mercedes and Audi were recorded. A ran

vladimir1956 [14]

Answer:

Median for sample 1 = 142

Median for sample 2 = 134

Step-by-step explanation:

For Sample 1:

For finding median, the data is first arranged into ascending or descending order. We are arranging the data in ascending order.

120, 130, 132, 136, 139, 142, 142, 144, 156, 165, 167

The formula for calculating the term which will be median is:

Median= ((n+1)/2)

Here in sample 1, n=11

So, putting n=11 in the formula

= ((11+1)/2)

= (12/2)

=6th term

The sixth term is 142, so

Median of sample 1=142

For Sample 2:

For finding median, the data is first arranged into ascending or descending order. We are arranging the data in ascending order.

112, 112, 113, 114, 125, 134, 141, 145, 146, 163, 165

The formula for calculating the term which will be median is:

Median= ((n+1)/2)

Here in sample 2, n=11

So, putting n=11 in the formula

=((11+1)/2)

=(12/2)

=6th term

The sixth term is 134, so

Median of sample 2=134

7 0

3 years ago

What is the length of the diagonal of a poster board with dimensions 22 inches by 28 inches? Round to the nearest tenth.

devlian [24]

Answer:

The length of the diagonal of a poster board is $35.6\ in$

Step-by-step explanation:

Let

x----> the length of the diagonal of a poster board

we know that

Applying the Pythagoras Theorem

$x^{2}=22^{2}+28^{2} \\ \\x^{2}=1,268\\ \\x=35.6\ in$

4 0

2 years ago

Read 2 more answers

There are three power plants [X, Y, Z] that at any given time each one either generates electricity or idles. Event A is that pl

insens350 [35]

We're told that

$P(A\cap B)=0.15$

$P(A\cup B)^C=0.06\implies P(A\cup B)=0.94$

$P(B\mid A)=P(B^C\mid A)=0.5$

where the last fact is due to the law of total probability:

$P(A)=P(A\cap B)+P(A\cap B^C)$

$\implies P(A)=P(B\mid A)P(A)+P(B^C\mid A)P(A)$

$\implies 1=P(B\mid A)+P(B^C\mid A)$

so that $B\mid A$ and $B^C\mid A$ are complementary.

By definition of conditional probability, we have

$P(B\mid A)=P(B^C\mid A)$

$\implies\dfrac{P(A\cap B)}{P(A)}=\dfrac{P(A\cap B^C)}{P(A)}$

$\implies P(A\cap B)=P(A\cap B^C)$

We make use of the addition rule and complementary probabilities to rewrite this as

$P(A\cap B)=P(A\cap B^C)$

$\implies P(A)+P(B)-P(A\cup B)=P(A)+P(B^C)-P(A\cup B^C)$

$\implies P(B)-[1-P(A\cup B)^C]=[1-P(B)]-P(A\cup B^C)$

$\implies2P(B)=2-[P(A\cup B)^C+P(A\cup B^C)]$

$\implies2P(B)=[1-P(A\cup B)^C]+[1-P(A\cup B^C)]$

$\implies2P(B)=P(A\cup B)+P(A\cup B^C)^C$

$\implies2P(B)=P(A\cup B)+P(A^C\cap B)\quad(*)$

By the law of total probability,

$P(B)=P(A\cap B)+P(A^C\cap B)$

$\implies P(A^C\cap B)=P(B)-P(A\cap B)$

and substituting this into $(*)$ gives

$2P(B)=P(A\cup B)+[P(B)-P(A\cap B)]$

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

$\implies P(B)=0.94-0.15=\boxed{0.79}$

8 0

2 years ago

How many pairs of parallel line does a triangle have

sammy [17]

Answer:

None.

Step-by-step explanation:

A triangle has 3 sides so any pair of parallel lines is impossible.

3 0

2 years ago

Read 2 more answers

Show all work and reasoning

Natalija [7]

Split up the interval [2, 5] into $n$ equally spaced subintervals, then consider the value of $f(x)$ at the right endpoint of each subinterval.

The length of the interval is $5-2=3$ , so the length of each subinterval would be $\dfrac3n$ . This means the first rectangle's height would be taken to be $x^2$ when $x=2+\dfrac3n$ , so that the height is $\left(2+\dfrac3n\right)^2$ , and its base would have length $\dfrac{3k}n$ . So the area under $x^2$ over the first subinterval is $\left(2+\dfrac3n\right)^2\dfrac3n$ .

Continuing in this fashion, the area under $x^2$ over the $k$ th subinterval is approximated by $\left(2+\dfrac{3k}n\right)^2\dfrac{3k}n$ , and so the Riemann approximation to the definite integral is

$\displaystyle\sum_{k=1}^n\left(2+\frac{3k}n\right)^2\frac{3k}n$

and its value is given exactly by taking $n\to\infty$ . So the answer is D (and the value of the integral is exactly 39).

8 0

3 years ago

20a-18 2: 42 3: 17b+3 4: 105