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3 years ago

A month of the year is chosen at random. What is the probability that the month starts with the letter Jor the letter M?

Mathematics

2 answers:

tamaranim1 [39]3 years ago

4 0

Answer:

5 out of 12

5/12

January, March, May, June, July

There are three months whose name starts with the letter 'j': January, June, July

There are two months whose name starts with the letter 'm': March, May

So altogether, there are 5 months whose name starts with the letter 'j' or 'm'. So the possible outcome is 5 and total number of sample space is 12.

Therefore the required probability is: 5/12

PLEASE GIVE BRAINLYEST :)

andrew-mc [135]3 years ago

4 0

Answer:

Step-by-step explanation:

Total outcomes = 12

Letter starting with M = 2 (march & May)

Letter starting with J = 3 ( January, June, July)

P(letter starting with M and J)=2/12+3/12=5/12

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Answer:

The break-even sales amounts is 36 or 224.

Step-by-step explanation:

Consider the provided function.

$P= -5x^2 + 13x - 4$

Where x is the number of televisions sold (in hundreds) and P is the profit.

We need to calculate the break-even sales amounts.

the break-even sales amounts is the sales amounts that result in no profit or loss.

That means substitute P=0 and solve for x.

$-5x^2 + 13x - 4=0$

$5x^2-13x+4=0$

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute a=4, b=-13 and c=4 in above formula.

$x_{1,\:2}=\frac{-\left(-13\right)\pm \sqrt{\left(-13\right)^2-4\cdot \:5\cdot \:4}}{2\cdot \:5}$

$x_{1,\:2}=\frac{13\pm\sqrt{169-80}}{10}$

$x_{1,\:2}=\frac{13\pm\sqrt{89}}{10}$

$x=\frac{13+\sqrt{89}}{10},\:x=\frac{13-\sqrt{89}}{10}\\x\approx2.243, \ or\ x\approx 0.357$

Therefore, the break-even sales amounts is 36 or 224.

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We are given the following hypothesis below;

Let p = <u><em>proportion of people who write with their left hand</em></u>

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