Multiply. Simplify answer. <br> 2*2 1/3

Anne wants to fill 12 hanging baskets with compost. Each hanging basket is a hemisphere of diameter 40 cm.

murzikaleks [220]

Answer:

no

Step-by-step explanation:

Steps to answering this question

determine the volume of the 12 basket

volume of a hemisphere = (2/3)πr^3

n = 22/7

r = radius

the diameter is the straight line that passes through the centre of a circle and touches the two edges of the circle.

A radius is half of the diameter

radius = 40/2 = 20 cm

volume of one hemisphere = (2/3) x (22/7) x (20^3) = 16,761.90 cm^3

volume of the 12 baskets = 16,761.90 cm^3 x 12 = 201,142.86 cm^

2. convert the litres of compost to cm and multiply by the total bags of compost

1 litre = 1000cm

1 bag of compost = 50 x 1000 = 50,000

4 bags of compost = 50,000 x 4 = 200,000 cm

3. compare which figure is higher. the figure gotten in step 1 or 2

201,142.86 cm^3 is greater than 200,000

there is no enough compost

473691.4

273691.4

3 0

3 years ago

Regroup hundreds as Tens:<br><br> 8 hundreds 9 tens =

Lilit [14]

89 tens because 8 x10 is 80 add the 9
89

7 0

3 years ago

For a certain river, suppose the drought length Y is the number of consecutive time intervals in which the water supply remains

AnnZ [28]

Answer:

a) There is a 9% probability that a drought lasts exactly 3 intervals.

There is an 85.5% probability that a drought lasts at most 3 intervals.

b)There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

Step-by-step explanation:

The geometric distribution is the number of failures expected before you get a success in a series of Bernoulli trials.

It has the following probability density formula:

$f(x) = (1-p)^{x}p$

In which p is the probability of a success.

The mean of the geometric distribution is given by the following formula:

$\mu = \frac{1-p}{p}$

The standard deviation of the geometric distribution is given by the following formula:

$\sigma = \sqrt{\frac{1-p}{p^{2}}$

In this problem, we have that:

$p = 0.383$

So

$\mu = \frac{1-p}{p} = \frac{1-0.383}{0.383} = 1.61$

$\sigma = \sqrt{\frac{1-p}{p^{2}}} = \sqrt{\frac{1-0.383}{(0.383)^{2}}} = 2.05$

(a) What is the probability that a drought lasts exactly 3 intervals?

This is $f(3)$

$f(x) = (1-p)^{x}p$

$f(3) = (1-0.383)^{3}*(0.383)$

$f(3) = 0.09$

There is a 9% probability that a drought lasts exactly 3 intervals.

At most 3 intervals?

This is $P = f(0) + f(1) + f(2) + f(3)$

$f(x) = (1-p)^{x}p$

$f(0) = (1-0.383)^{0}*(0.383) = 0.383$

$f(1) = (1-0.383)^{1}*(0.383) = 0.236$

$f(2) = (1-0.383)^{2}*(0.383) = 0.146$

Previously in this exercise, we found that $f(3) = 0.09$

So

$P = f(0) + f(1) + f(2) + f(3) = 0.383 + 0.236 + 0.146 + 0.09 = 0.855$

There is an 85.5% probability that a drought lasts at most 3 intervals.

(b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

This is $P(X \geq \mu+\sigma) = P(X \geq 1.61 + 2.05) = P(X \geq 3.66) = P(X \geq 4)$ .

We are working with discrete data, so 3.66 is rounded up to 4.

Either a drought lasts at least four months, or it lasts at most thee. In a), we found that the probability that it lasts at most 3 months is 0.855. The sum of these probabilities is decimal 1. So:

$P(X \leq 3) + P(X \geq 4) = 1$

$0.855 + P(X \geq 4) = 1$

$P(X \geq 4) = 0.145$

There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

8 0

3 years ago

65 is 20% of what number

Verizon [17]

(60 / - 20 ) x 100% = 300%

6 0

3 years ago

Noel has rowing lessons every 5 days and guitar lessons every 6 days. If he had both lessons on the last day of the previous mon

bixtya [17]

Answer:

Since every 30 days he wil have both lessons on the same day , and he already had both lessons on the last day of the previous month, that means that the day 30 the current month he wil have both lessons on the same day (It may be the last day if the month has 30 days or it may not be the last day if the month has 31 days)

Step-by-step explanation:

Lets find the least common factor of 5 and 6

Multiples of 5

5 10 15 20 35 30 35 40......

Multiples of 6

6 12 18 24 30 36

LCF of 5 and 6 = 30

Every 30 days he wil have both lessons on the same day

3 0

3 years ago

Multiply. Simplify answer. 2*2 1/3