What is midpoint 85 90?

The speeds of cars on a given i street we are normally distributed with a mean of 72 miles per hour and a standard deviation of

Ostrovityanka [42]

Answer: 72.78% of the drivers are traveling between 70 and 80 miles per hour based on this distribution.

Step-by-step explanation:

Let X be a random variable that represents the speed of the drivers.

Given: population mean : M = 72 miles ,

Standard deviation: s= 3.2 miles

The probability that the drivers are traveling between 70 and 80 miles per hour based on this distribution:

$P(70\leq X\leq 80)=P(\frac{70-72}{3.2}\leq \frac{X-M}{s}\leq\frac{80-72}{3.2})\\\\= P(-0.625\leq Z\leq 2.5)\ \ \ \ \ [Z=\frac{X-M}{s}]\\\\=P(Z\leq2.5)-P(Z\leq -0.625)\\\\\\ =0.9938-0.2660\ \ \ [\text{Using p-value calculator}]\\\\=0.7278$

Hence, 72.78% of the drivers are traveling between 70 and 80 miles per hour based on this distribution.

3 0

3 years ago

What is the greatest common factor GCF of 10 and four

Pani-rosa [81]

Answer:

Take the numbers 50 and 30. Their greatest common factor is 10, since 10 is the greatest factor that both numbers have in common. To find the GCF of greater numbers, you can factor each number to find their prime factors, identify the prime factors they have in common, and then multiply those together.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

I’m having trouble with this

Alchen [17]

Answer:

Step-by-step explanation:

EVEN THOUGH IT SHOWS A 1 DEGREE TEMPERATURE RISE AT 4PM TO 66

THAT SHOULD NOT MATTER IT FELL A TOTAL OF 5 DEGREES FROM 3 TO 6PM

4 0

3 years ago

Suppose that in a certain county 37% of voters are registered as Democrats, 29% as Republicans, 7% as Green party, and the rest

Fantom [35]

Answer:

$0.66$

Step-by-step explanation:

GIVEN: Suppose that in a certain county $37\%$ of voters are registered as Democrats, $29\%$ as Republicans, $7\%$ as Green party, and the rest are considered Independents. You conduct a poll by calling registered voters in the county at random.

TO FIND: Probability that the first call will be to either a Democrat or a Republican.

SOLUTION:

Lets total population of county be $100x$ .

voters registered as Democrats $=37\%=\frac{37}{100}\times100x$

$=37x$

voters registered as Republicans $=29\%=\frac{29}{100}\times100x$

$=29x$

voters registered as Green party $=7\%=\frac{7}{100}\times100x$

$=7x$

Probability that first call will be to Democrats $P(A)=\frac{\text{total democrats}}{\text{total voters}}$

$=\frac{37x}{100x}=0.37$

Probability that first call will be to Republican $P(B)=\frac{\text{total republicans}}{\text{total voters}}$

$=\frac{29x}{100x}=0.29$

probability that the first call will be to either a Democrat or a Republican

$=P(A)+P(B)$

$=0.37+0.29$

$=0.66$

Hence the probability that the first call will be to either a Democrat or a Republican is $0.66$

6 0

3 years ago

A fair die is rolled 12 times. the number of times an even number occurs on the 12 rolls has

bonufazy [111]

Answer:

Step-by-step explanation:

For a fair die, there are six likely options; 1, 2, 3, 4, 5, and 6

the probability of a even number is 3/6 = 0.5

Since the results of the die roll is independent and each trial is mutually exclusive, the distribution to explain the probability of occurrence will follow a binomial distribution such that n is the number of trials

x = number of successful throws

therefore for a Binomial distribution where

P(X =x) = nCx . P^x . (1-P)^ (n-x)

since p = 0.5, and n = 12, the distribution follows

P(X = x) = 12Cx . 0.5^x . (1 - 0.5)^(12- x)

= 12Cx . 0.5^x . 0.5)^(12- x)

where x = (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12)

since we are interested in the probability of the number of times an even number occurs

it can occur either as P(X = 0), P(X =1), P(X =2), P(X =3), P(X =4), P(X =5), P(X =6), P(X =7), P(X =8), P(X =9), P(X =10), P(X =11), and P(X =12)

For no even number in 12 rolls,

P(X = 0) = 12C0 . 0.5^0 . 0.5^(12- 0) = 0.000244

For one even number in 12 rolls,

P(X = 1) = 12C1 . 0.5^1 . 0.5^(12- 1) = 0.002930

For two even number in 12 rolls,

P(X = 2) = 12C2 . 0.5^2 . 0.5^(12- 2) = 0.016113

For three even number in 12 rolls,

P(X = 3) = 12C3 . 0.5^3 . 0.5^(12- 3) = 0.053711

For four even number in 12 rolls,

P(X = 4) = 12C4 . 0.5^4 . 0.5^(12- 4) = 0.120850

For five even number in 12 rolls,

P(X = 5) = 12C5 . 0.5^5 . 0.5^(12- 5) = 0.193359

For six even number in 12 rolls,

P(X = 6) = 12C6 . 0.5^6 . 0.5^(12- 6) = 0.225586

For seven even number in 12 rolls,

P(X = 7) = 12C7 . 0.5^7 . 0.5^(12- 7) = 0.193359

For eight even number in 12 rolls,

P(X = 8) = 12C8 . 0.5^8 . 0.5^(12- 8) = 0.120850

For nine even number in 12 rolls,

P(X = 9) = 12C9 . 0.5^9 . 0.5^(12- 9) = 0.053711

For ten even number in 12 rolls,

P(X = 10) = 12C10 . 0.5^10 . 0.5^(12- 10) = 0.016113

For eleven even number in 12 rolls,

P(X = 11) = 12C11 . 0.5^11 . 0.5^(12- 11) = 0.002930

For twelve even number in 12 rolls,

P(X = 12) = 12C12 . 0.5^12 . 0.5^(12- 12) = 0.000244

Final test summation[P(X)] = 1

i.e.

P(X = 0) + P(X =1) + P(X =2) + P(X =3) + P(X =4) + P(X =5) + P(X =6) + P(X =7) + P(X =8) + P(X =9) + P(X =10) + P(X =11) + P(X =12) = 1

Hence since 0.000244 + 0.002930 + 0.016113 + 0.053711 + 0.120850 + 0.193359 + 0.225586 + 0.193359 + 0.120850 + 0.053711 + 0.016113 + 0.002930 + 0.000244 = 1.000000,

the probability value stands

7 0

3 years ago