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KATRIN_1 [288]
2 years ago
6

PLease help meeeeeeeee!

Mathematics
2 answers:
pentagon [3]2 years ago
5 0

Answer: I don't know

Step-by-step explanation:

ValentinkaMS [17]2 years ago
4 0

The photo is not clear.......

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Umnica [9.8K]

Answer:

It b

Step-by-step explanation:

We know this because 1/2 is 0.5 while 2/3 is 0.67. With this we can say that 1/2 is least than 2/3

Hope this Helped

7 0
2 years ago
1
melamori03 [73]

Answer: 162.00 or 40.50 or 202.50

Step-by-step explanation:

$6.75 times 30 = 202.50

202.50 times 0.2 = 40.50

202.50 minus 40.50 = 162.00

7 0
3 years ago
How do I find x (area of rectangle & trapezium?
lesantik [10]
Rectangle Formula: a = lw (Area = Length x Width) 

Trapezium: A= a+b/2h (Area = Area + Base / 2 x h) 

I hope this helps. :)
5 0
3 years ago
What is the equation in slope-intercept from of the line that crosses the x-axis at 16 and is perpendicular to the line represen
Katen [24]

Answer:

35

Step-by-step explanation:

8 0
3 years ago
In a certain population of the eastern thwump bird, the wingspan of the individual birds follows an approximately normal distrib
Greeley [361]

Answer:

a) P(48 < x < 58) = 0.576

b) P(X ≥ 1) = 0.9863

c) E(X) 2.88

d) P(x < 48) = 0.212

e) P(X > 2) = 0.06755

Step-by-step explanation:

The mean of the wingspan of the birds = μ = 53.0 mm

The standard deviation = σ = 6.25 mm

a) Probability of a bird having a wingspan between 48 mm and 58 mm can be found by modelling the problem as a normal distribution problem.

To solve this, we first normalize/standardize the two wingspans concerned.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ

For wingspan 48 mm

z = (48 - 53)/6.25 = - 0.80

For wingspan 58 mm

z = (58 - 53)/6.25 = 0.80

To determine the probability that the wingspan of the first bird chosen is between 48 and 58 mm long. P(48 < x < 58) = P(-0.80 < z < 0.80)

We'll use data from the normal probability table for these probabilities

P(48 < x < 58) = P(-0.80 < z < 0.80) = P(z < 0.8) - P(z < -0.8) = 0.788 - 0.212 = 0.576

b) The probability that at least one of the five birds has a wingspan between 48 and 58 mm = 1 - (Probability that none of the five birds has a wingspan between 48 and 58 mm)

P(X ≥ 1) = 1 - P(X=0)

Probability that none of the five birds have a wingspan between 48 and 58 mm is a binomial distribution problem.

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = number of birds = 5

x = Number of successes required = number of birds with wingspan between 48 mm and 58 mm = 0

p = probability of success = Probability of one bird having wingspan between 48 mm and 58 mm = 0.576

q = probability of failure = Probability of one bird not having wingspan between 48 mm and 58 mm = 1 - 0.576 = 0.424

P(X=0) = ⁵C₀ (0.576)⁰ (1 - 0.576)⁵ = (1) (1) (0.424)⁵ = 0.0137

The probability that at least one of the five birds has a wingspan between 48 and 58 mm = P(X≥1) = 1 - P(X=0) = 1 - 0.0137 = 0.9863

c) The expected number of birds in this sample whose wingspan is between 48 and 58 mm.

Expected value is a sum of each variable and its probability,

E(X) = mean = np = 5×0.576 = 2.88

d) The probability that the wingspan of a randomly chosen bird is less than 48 mm long

Using the normal distribution tables again

P(x < 48) = P(z < -0.8) = 1 - P(z ≥ -0.8) = 1 - P(z ≤ 0.8) = 1 - 0.788 = 0.212

e) The probability that more than two of the five birds have wingspans less than 48 mm long = P(X > 2) = P(X=3) + P(X=4) + P(X=5)

This is also a binomial distribution problem,

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = number of birds = 5

x = Number of successes required = number of birds with wingspan less than 48 mm = more than 2 i.e. 3,4 and 5.

p = probability of success = Probability of one bird having wingspan less than 48 mm = 0.212

q = probability of failure = Probability of one bird not having wingspan less than 48 mm = 1 - p = 0.788

P(X > 2) = P(X=3) + P(X=4) + P(X=5)

P(X > 2) = 0.05916433913 + 0.00795865476 + 0.00042823218

P(X > 2) = 0.06755122607 = 0.06755

5 0
3 years ago
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