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aleksandrvk [35]
3 years ago
8

Is -134 greater than or less than 1136

Mathematics
1 answer:
masya89 [10]3 years ago
7 0

Answer:

less than.

Step-by-step explanation:

.............

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For 3/4 of a pound of peanuts, Jimmy paid $2.35. What is the unit rate (dollars per pound)? *
Tpy6a [65]

Answer:

$3.13/lb

Step-by-step explanation:

Divide 3/4 pound into $2.35:

$2.35

--------- = $3.13/lb

3/4  lb

8 0
2 years ago
Huddson has practiced a 360-back flip for months. This week, he did it perfectly 24 out of 54 times. At this rate, how many time
iris [78.8K]
Let x = perfect flips out of nine
\frac{24}{54}  =  \frac{x}{9}
54x = 216
x = 4
4 flips
5 0
3 years ago
HELP ME PLEASE
aivan3 [116]

Answer:

Answer is A

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
Need help on number 2 plz I appreciate it thank you so much
vladimir1956 [14]

Answer:

\angle PQR, \angle SQR, and \angle PQS

Or

angle PQR, angle SQR and angle PQS

Step-by-step explanation:

The three different angles in the diagram are angle PQR, angle SQR and angle PQS.

Another way of writing this is using an angle sign before the alphabets follows. Thus:

\angle PQR, \angle SQR, and \angle PQS

6 0
3 years ago
Let X denote the distance (m) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose t
andrezito [222]

Answer and Step-by-step explanation: For an exponential distribution, the probability distribution function is:

f(x) = λ.e^{-\lambda.x}

and the cumulative distribution function, which describes the probability distribution of a random variable X, is:

F(x) = 1 - e^{-\lambda.x}

(a) <u>Probability</u> of distance at most <u>100m</u>, with λ = 0.0143:

F(100) = 1 - e^{-0.0143.100}

F(100) = 0.76

<u>Probability</u> of distance at most <u>200</u>:

F(200) = 1 - e^{-0.0143.200}

F(200) = 0.94

<u>Probability</u> of distance between <u>100 and 200</u>:

F(100≤X≤200) = F(200) - F(100)

F(100≤X≤200) = 0.94 - 0.76

F(100≤X≤200) = 0.18

(b) The mean, E(X), of a probability distribution is calculated by:

E(X) = \frac{1}{\lambda}

E(X) = \frac{1}{0.0143}

E(X) = 69.93

The standard deviation is the square root of variance,V(X), which is calculated by:

σ = \sqrt{\frac{1}{\lambda^{2}} }

σ = \sqrt{\frac{1}{0.0143^{2}} }

σ = 69.93

<u>Distance exceeds the mean distance by more than 2σ</u>:

P(X > 69.93+2.69.93) = P(X > 209.79)

P(X > 209.79) = 1 - P(X≤209.79)

P(X > 209.79) = 1 - F(209.79)

P(X > 209.79) = 1 - (1 - e^{-0.0143*209.79})

P(X > 209.79) = 0.0503

(c) Median is a point that divides the value in half. For a probability distribution:

P(X≤m) = 0.5

\int\limits^m_0 f({x}) \, dx = 0.5

\int\limits^m_0 {\lambda.e^{-\lambda.x}} \, dx = 0.5

\lambda.\frac{e^{-\lambda.x}}{-\lambda} = -e^{-\lambda.x} + e^{0}

1 - e^{-\lambda.m} = 0.5

-e^{-\lambda.m} = - 0.5

ln(e^{-0.0143.m}) = ln(0.5)

-0.0143.m = - 0.0693

m = 48.46

6 0
3 years ago
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