A. T^26<br> B. T^-26<br> C. T^38<br> D. T^-38

Is this right I'm practicing multiplying decimals by 10, 100, or 1,000

Leokris [45]

All are correct except the bottom, it would just be 4.23 since there are no zeroes, any number times 1 is itself

8 0

3 years ago

Round 0.832 to the nearest hundredth.

kodGreya [7K]

Answer:

0.83

Step-by-step explanation:

5 or more raise the score 4 or less let it rest

Elementary nostoligia :)

3 0

3 years ago

Read 2 more answers

Suppose that the probability of a drought in any independent year is 20%. Out of those years in which a drought occurs, the prob

astraxan [27]

Answer:

the correct option is d. 0.02

Step-by-step explanation:

We have the following events:

x: A drought occurd

y: Water rationing.

Therefore, according to the plaster we have:

P [x] = 0.20

P [y | x] = 0.10

Now P [it is a drought and water rationing happens] =

P [x n y] = P [y | x] * P [x] = 0.10 * 0.20 = 0.02

Which means that the correct option is d. 0.02

3 0

3 years ago

Consider a uniform distribution from aequals4 to bequals29. (a) Find the probability that x lies between 7 and 27. (b) Find th

weeeeeb [17]

Answer:

a) 80% probability that x lies between 7 and 27.

b) 28% probability that x lies between 6 and 13.

c) 44% probability that x lies between 9 and 20.

d) 28% probability that x lies between 11 and 18.

Step-by-step explanation:

An uniform probability is a case of probability in which each outcome is equally as likely.

For this situation, we have a lower limit of the distribution that we call a and an upper limit that we call b.

The probability that we find a value x between c and d, in which d is larger than c, is given by the following formula.

$P(c \leq x \leq d) = \frac{d - c}{b - a}$

Uniform distribution from a = 4 to b = 29

(a) Find the probability that x lies between 7 and 27.

So $c = 7, d = 27$

$P(7 \leq x \leq 27) = \frac{27 - 7}{29 - 4} = 0.8$

80% probability that x lies between 7 and 27.

(b) Find the probability that x lies between 6 and 13. 

So $c = 6, d = 13$

$P(6 \leq x \leq 13) = \frac{13 - 6}{29 - 4} = 0.28$

28% probability that x lies between 6 and 13.

(c) Find the probability that x lies between 9 and 20.

So $c = 9, d = 20$

$P(9 \leq x \leq 20) = \frac{20 - 9}{29 - 4} = 0.44$

44% probability that x lies between 9 and 20.

(d) Find the probability that x lies between 11 and 18.

So $c = 11, d = 18$

$P(11 \leq x \leq 18) = \frac{18 - 11}{29 - 4} = 0.28$

28% probability that x lies between 11 and 18.

3 0

3 years ago

Can someone please help me with this problem!

Anna71 [15]

Answer:

11 yards

Step-by-step explanation:

The volume of a cone formula is:

$V=\frac{1}{3}\pi r^2 h$

Where V is the volume,

r is the radius (half of diameter)

h is the height

Given,

h = 15

V = 475.17

We find the radius (r) first by substituting given values:

$V=\frac{1}{3}\pi r^2 h\\475.17=\frac{1}{3}\pi r^2 (15)\\475.17=5\pi r^2\\r^2=30.25\\r=5.5$

The radius (r) is 5.5 yards

We need the diameter, which is DOUBLE THE RADIUS, so diameter would be:

Diameter = 5.5 * 2 = 11 yards

8 0

3 years ago

A. T^26 B. T^-26 C. T^38 D. T^-38