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2 years ago

14

Find the surface area of the solid figure represented by the given net. 10cm squared 10cm squared 10cm squared 10cm squared 25cm

squared

2 answers:

m_a_m_a [10]2 years ago

7 0

Answer:

its actually 65

Step-by-step explanation:

Vika [28.1K]2 years ago

5 0

GIVEN:

Side = 5cm.

FORMULA:

Surface area of cube = 6side²

ANSWER:

We know the formula,

so SA = 6 × (5)²

6 × 25
150cm².

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59.7

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Jeff hiked for 2 hours and traveled 5 miles. If he continues at the same pace, which equation will show the relationship between

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Read 2 more answers

Data collected at Toronto Pearson International Airport suggests that an exponential distribution with mean value 2725hours is a

Ivan

Answer:

a) What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

$P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48$

At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

b) What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

$P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X$

What is the probability that it is less than the mean value by more than one standard deviation?

$P(X$

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

$P(X=x)=\lambda e^{-\lambda x}$

The cumulative distribution for this function is given by:

$F(X) = 1- e^{-\lambda x}, x\ geq 0$

We know the value for the mean on this case we have that :

$mean = \frac{1}{\lambda}$

$\lambda = \frac{1}{Mean}= \frac{1}{2.725}=0.367$

Solution to the problem

Part a

What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

$P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48$

At most 3 hours?

$P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667$

Part b

What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

The variance for the esponential distribution is given by: $Var(X) =\frac{1}{\lambda^2}$

And the deviation would be:

$Sd(X) = \frac{1}{\lambda}= \frac{1}{0.367}= 2.725$

And the mean is given by $Mean = 2.725$

Two deviations correspond to 5.540, so we want this probability:

$P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X$

What is the probability that it is less than the mean value by more than one standard deviation?

For this case we want this probablity:

$P(X$

8 0

2 years ago

Help please and ill mark you brainliest :)

rodikova [14]

Answer:

n=100+0.05(460)

Step-by-step explanation:

She earns 100 dollars all the time. so its 100+ something

she also earns 5% of the tips, which in that case were 460$

you can repersent that as 0.05(460)

so: 100+0.05(460)

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2 years ago

9. If each of the following pairs represents a pair of equivalent rational

Oksi-84 [34.3K]

Answer:

-35/6

Step-by-step explanation:

We need to find the value of x , And the given pair of Rational numbers are equal.

<u>Question</u><u> </u><u>(</u><u>I)</u><u> </u><u>:</u><u>-</u><u> </u>

→ 3/4 = 7/x

→ 7/x = 3/4

→ 1/x = 3/4 × 1/7

→ 1/x = 3/28

→ x = 28/3

<u>Question</u><u> (</u><u>ii)</u><u> </u><u>:</u><u>-</u><u> </u>

→ x/7 = -5/6

→ x = -5/6 × 7

→ x = -35/6

<h3>• Hence the value of x is -35/6 .</h3>

7 0

2 years ago