2 dot plots. The highlands have a mean rainfall of 15.27 millimeters, and the Lowlands have a mean rainfall of 12.05 millimeters
.
The dot plots show rainfall totals for several spring storms in highland areas and lowland areas.
What is the mean rainfall for the highland storms?
What is the mean rainfall for the lowland storms?
The dot plots show rainfall totals for several spring storms in highland areas and lowland areas.
What is the mean rainfall for the highland storms?
What is the mean rainfall for the lowland storms?
2 answers:
The dot plots show rainfall totals for several spring storms in highland areas and lowland areas.
What is the mean rainfall for the highland storms?
✔ 15.27 mm
What is the mean rainfall for the lowland storms?
✔ 12.05 mm
can i plz have brainley i just did the answer and got it right
and do you like my drawing i just made is on my laptop (its kinda hard).
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Answer: $ 4,515
Step-by-step explanation:
The cost of the car she wanted to buy = $35,550
Her savings = $20,700
That means she needs( $35,550 - $20,700) more in order for her to buy the car.
Amount borrowed from the finance company = $35,550 - $20,700
= $ 14,850
Since she now owes the financial company a total of $19,365, then the interest accrued by the loan she borrowed is given by:
$19,365 - $ 14,850
= $4,515
Answer:
The probabilities of each outcome are the following:
for X = 1 is 0.6
for X = 2 is 0.35
for X = 3 is 0.049
and for X = 4 is 0.001
Step-by-step explanation:
Let's consider X as the random variable for the sum of outcomes "S" exceeds 3, this is:
Let's now analyze and consider the ways that X equals the different values:
X = 1: I throw the dice and the result is: 4, 5 and 6.
Then: P(X=1) = P(4) + P(5) + P(6) = 0.2+0.1+0.3 = 0.6
Thus P(X=1) = 0.6
X = 2: The result after following the dice two times can be:
1 and 3, 1 an 4.... and so on until 1 and 6
2 and 2, 2 and 3... and so on until 2 and 6
3 and 1... and so on until 3 a 6
Then P(X=2) = P(1)xP(3,4,5,6) + P(2)xP(2....6) + P(3)x(1.....6)
Theory of Probability: Sum of all possible outcomes P(1)+P(2).......P(6) = 1
Then P(1......6) = 1
Then P(X=2) = P(1)x[1-P(1)-P(2)]+P(2)x[1-P(1)]+P(3) = 0.1x(1-0.1-0.2) + 0.2x(1-0.1) + 0.1 = 0.1 x 0.7 + 0.2 x 0.9 + 0.1 = 0.35
Thus P(X=2) = 0.35
X = 3: The result can be
1 and 1 and 2, 1 and 1 and 3.... until 1 and 1 and 6
1 and 2 and 1, 1 and 2 and 2..... until 1 and 2 and 6
2 and 1 and 1, 2 and 1 and 2.... until 2 and 1 and 6
Then P(X=3) = P(1)xP(1)xP(2....6) + P(1)xP(2)xP(1.....6) + P(1)xP(2)xP(1.....6)
P(X=3) = 0.1 x 0.1 x (1-0.1) + 0.1 x 0.2 x 1 + 0.2 x 0.1 x 1 = 0.01 x 0.9 + 0.2 + 0.2 = 0.049
Thus P(X=3) = 0.049
Finally, for X to be 4, I only have the following possibilities
1 and 1 and 1 and 1.... until 1 and 1 and 1 and 6
Then P(X=4) = P(1)xP(1)xP(1)xP(1.....6) = 0.1x0.1x0.1x1 = 0.001
Thus P(X=3) = 0.001