Is -134 greater than or less than 1136

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) Probability of distance at most 100m, with λ = 0.0143:

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

F(100) = 0.76

Probability of distance at most 200:

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

F(200) = 0.94

Probability of distance between 100 and 200:

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

Distance exceeds the mean distance by more than 2σ:

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