Find the positive solution for c: 8 2 + 15 2 = c 2

Harold, Clara, and Mark each chose a different lunch from three categories:

Yanka [14]

Answer:

a) Harold

b) Clara

Step-by-step explanation:

Each person has chose a different lunch. Based on this context, it means no one can have the same lunch item.

A explanation)

Mark must eat soup since he doesn't eat bread (sandwich) or salads.
Clara must eat salad since Mark is eating the soup and Clara doesn't eat sandwiches.
Therefore, Harold must eat the only different lunch item; sandwich. Since Harold had the sandwich, he also has the milk.

B explanation)

The person who had soup had iced tea; meaning Mark drank Iced Tea.
The person who had the sandwich also had the milk; Harold drank milk.
The only person missing a drink is Clara, so she ordered mineral water.

8 0

3 years ago

If XX is a binomial random variable, compute the mean, the standard deviation, and the variance for each of the following cases:

Liula [17]

Answer:

(a) The mean, variance and standard deviation of X are 1.60, 0.96 and 0.98 respectively.

(b) The mean, variance and standard deviation of X are 3.20, 0.64 and 0.80 respectively.

(c) The mean, variance and standard deviation of X are 1.50, 0.75 and 0.87 respectively.

(d) The mean, variance and standard deviation of X are 4.00, 0.80 and 0.89 respectively.

Step-by-step explanation:

The random variable X follows a Binomial distribution with parameters n and p.

The mean, variance and standard deviation of X are:

$\mu=np\\\sigma^{2}=np(1-p)\\\sigma=\sqrt{np(1-p)}$

(a)

For n = 4 and p = 0.40 compute the mean, variance and standard deviation of X as follows:

$\mu=np=4\times0.40=1.60\\\sigma^{2}=np(1-p)=4\times0.40\times(1-0.40)=0.96\\\sigma=\sqrt{np(1-p)}=\sqrt{0.96}=0.98$

Thus, the mean, variance and standard deviation of X are 1.60, 0.96 and 0.98 respectively.

(b)

For n = 4 and p = 0.80 compute the mean, variance and standard deviation of X as follows:

$\mu=np=4\times0.80=3.20\\\sigma^{2}=np(1-p)=4\times0.80\times(1-0.80)=0.64\\\sigma=\sqrt{np(1-p)}=\sqrt{0.64}=0.80$

Thus, the mean, variance and standard deviation of X are 3.20, 0.64 and 0.80 respectively.

(c)

For n = 3 and p = 0.50 compute the mean, variance and standard deviation of X as follows:

$\mu=np=3\times0.50=1.50\\\sigma^{2}=np(1-p)=3\times0.50\times(1-0.50)=0.75\\\sigma=\sqrt{np(1-p)}=\sqrt{0.75}=0.87$

Thus, the mean, variance and standard deviation of X are 1.50, 0.75 and 0.87 respectively.

(d)

For n = 5 and p = 0.80 compute the mean, variance and standard deviation of X as follows:

$\mu=np=5\times0.80=4.00\\\sigma^{2}=np(1-p)=5\times0.80\times(1-0.80)=0.80\\\sigma=\sqrt{np(1-p)}=\sqrt{0.80}=0.89$

Thus, the mean, variance and standard deviation of X are 4.00, 0.80 and 0.89 respectively.

6 0

3 years ago

Is 85 a multiple of 10?

blagie [28]

No, only multiples of 10 end in a 0

hope this helps :)

4 0

3 years ago

Read 2 more answers

The value of 2 in in 2783 is. times the value of 2 in 7283

nikitadnepr [17]

The value of 2 in 2,783 is 2,000

The value of 2 in 7,283 is 200

2,000 x 200 = 400,000

Hope this helps

-AaronWiseIsBae

3 0

3 years ago

Solve the equation. 5+4x-7=4x-2-x

larisa86 [58]

Answer
4x-4x+x=-2+7-5
x=0

3 0

3 years ago