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3 years ago

If a leaf is 8 centimetres long. About (to the nearest whole number) how long is the leaf in inches?

Mathematics

2 answers:

Andrej [43]3 years ago

7 0

Answer:

Step-by-step explanation:

just simple

Dmitry_Shevchenko [17]3 years ago

3 0

Answer:

3 inches rounded but actually 3.14961

Step-by-step explanation:

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A skydiving company insists

34kurt

Answer:

A skydiving company insists that its customers weigh at least 130 pounds, but no more than 280 pounds, including parachute and other gear. If the total weight of all gear is 25 pounds, write and solve a compound inequality that represents the weight range without gear that is acceptable.

105 ≤ x ≤ 255

think this is what you meant . . .

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3 years ago

Cassie has 1/2 pound of sugar in her cabinet.Her cake recipe calls

BigorU [14]

Answer:

2 cakes with the sugar she has and there will still be some sugar left.

Step-by-step explanation:

Amount of sugar in the cabinet of Cassie = 1/2 pound

Amount of sugar required by Cassie to make her cake recipe = 2/10 pound

= 1/5 pounds

Then

The number of cakes that Cassie can make = (1/2)/(1/5)

= 5/2

= 2 1/2

6 0

3 years ago

What is the equivalent of | -21 |

Fynjy0 [20]

That's 21.

Hope this helps !

Nyway

8 0

3 years ago

Read 2 more answers

Indicate whether or not 1/√2 belongs to each set of numbers

Oksi-84 [34.3K]

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Hey mate Whole numbers is your Answer

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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

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