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3 years ago

How do you find range

Mathematics

2 answers:

Tamiku [17]3 years ago

5 0

First order the data from least to greatest, then subtract the smallest value from the largest value in the set.

Ivanshal [37]3 years ago

5 0

Order the numbers from least to greatest. Then subtract the Greater number by the Smaller number.

Example:
4, 6, 2, 3, 4, 10, 4, 7
Step One- Order from least to greatest: 2, 3, 4, 4, 4, 7, 10
Step Two- Subtract the Greatest by the Least: 10-2
Answer would be 6.
(This is just an example. Don't use on answer unless it asks for an example)

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If one bottle of soda costs .99, how much do three bottles cost? how much do they cost with 5% tax added on?

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3 years ago

Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive

LenaWriter [7]

Answer:

(a) The probability that all the next three vehicles inspected pass the inspection is 0.343.

(b) The probability that at least 1 of the next three vehicles inspected fail is 0.657.

(d) The probability that at most 1 of the next three vehicles passes is 0.216.

(e) The probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

Step-by-step explanation:

Let X = number of vehicles that pass the inspection.

The probability of the random variable X is P (X) = 0.70.

(a)

Compute the probability that all the next three vehicles inspected pass the inspection as follows:

P (All 3 vehicles pass) = [P (X)]³

$=(0.70)^{3}\\=0.343$

Thus, the probability that all the next three vehicles inspected pass the inspection is 0.343.

(b)

Compute the probability that at least 1 of the next three vehicles inspected fail as follows:

P (At least 1 of 3 fails) = 1 - P (All 3 vehicles pass)

$=1-0.343\\=0.657$

Thus, the probability that at least 1 of the next three vehicles inspected fail is 0.657.

(c)

Compute the probability that exactly 1 of the next three vehicles passes as follows:

P (Exactly one) = P (1st vehicle or 2nd vehicle or 3 vehicle)

= P (Only 1st vehicle passes) + P (Only 2nd vehicle passes)

+ P (Only 3rd vehicle passes)

$=(0.70\times0.30\times0.30) + (0.30\times0.70\times0.30)+(0.30\times0.30\times0.70)\\=0.189$

Thus, the probability that exactly 1 of the next three vehicles passes is 0.189.

(d)

Compute the probability that at most 1 of the next three vehicles passes as follows:

P (At most 1 vehicle passes) = P (Exactly 1 vehicles passes)

+ P (0 vehicles passes)

$=0.189+(0.30\times0.30\times0.30)\\=0.216$

Thus, the probability that at most 1 of the next three vehicles passes is 0.216.

(e)

Let X = all 3 vehicle passes and Y = at least 1 vehicle passes.

Compute the conditional probability that all 3 vehicle passes given that at least 1 vehicle passes as follows:

$P(X|Y)=\frac{P(X\cap Y)}{P(Y)} =\frac{P(X)}{P(Y)} =\frac{(0.70)^{3}}{[1-(0.30)^{3}]} =0.3525$

Thus, the probability that all 3 vehicle passes given that at least 1 vehicle passes is 0.3525.

7 0

3 years ago

Graph a triangle (STU) and reflect it over the y-axis to create triangle ST'U'.

lukranit [14]

The x-coordinates of $\triangle S'T'U'$ will be the negation of the x-coordinates of $\triangle STU$

The line segment from S to the y-axis equals the line segment from S' to the y-axis. Similarly, the line segment from T to the y-axis equals the line segment from T' to the y-axis

See attachment for $\triangle STU$ and $\triangle S'T'U'$

In order to solve this question, I will make the following assumptions.

Assume that the coordinates of $\triangle STU$ are

$S = (4,5)$