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Harrizon [31]
3 years ago
13

135 pizzas cost $1,405, how much does just 1 pizza cost?

Mathematics
2 answers:
Delicious77 [7]3 years ago
6 0

Answer: $10.41

Step-by-step explanation: 1405/135 = 10.407, so rounded is $10.41 since you round to the nearest hundredth when dealing with money.

Hope this helps!

Mark Brainliest if you want!

frez [133]3 years ago
5 0

Answer:

l10.4074074074

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thirty percent of all the students in a school are in a play. all students exept for 140 are in the play. how many students are
iogann1982 [59]

There is exactly 200 students in the school.

In order to find this, we know that 70% of students are not in the play (because 30% was in the play). Now, in order to solve for the total number, we can divide the students not in the play by the 70%.

140/.70 = 200 Students.

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Write a equation in slope intercept form that is equivalent to 10x-2y=16
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Answer:

y = 5x - 8

Step-by-step explanation:

10x -16 = 2y

y = 10/2x - 16/2

y = 5x - 8

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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 <em>X </em>are 1.60, 0.96 and 0.98 respectively.

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

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

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

Step-by-step explanation:

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The mean, variance and standard deviation of <em>X</em> are:

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

(a)

For <em>n</em> = 4 and <em>p</em> = 0.40 compute the mean, variance and standard deviation of <em>X </em>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 <em>X </em>are 1.60, 0.96 and 0.98 respectively.

(b)

For <em>n</em> = 4 and <em>p</em> = 0.80 compute the mean, variance and standard deviation of <em>X </em>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

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(c)

For <em>n</em> = 3 and <em>p</em> = 0.50 compute the mean, variance and standard deviation of <em>X </em>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

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(d)

For <em>n</em> = 5 and <em>p</em> = 0.80 compute the mean, variance and standard deviation of <em>X </em>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 <em>X </em>are 4.00, 0.80 and 0.89 respectively.

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3 years ago
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Answer:

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I hope this helps!

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