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3 years ago

What is 1/6 plus 3/8 all together

1 answer:

7 0

$\frac{1}{6} + \frac{3}{8} = \frac{1\times 4}{6\times 4} + \frac{3\times 3}{8\times 3} = \frac{4}{24} + \frac{9}{24} = \frac{9+4}{24} = \boxed{\frac{13}{24}}}$

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Which is the upper left quadrant on the coordinate plane?

Lisa [10]

Answer:

the upper left Quadrant is Quadrant II

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3 years ago

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Kaitlyn's older brother borrowed money for school he took out a loan that charges 6% simple interest he will end up paying 720 a

tankabanditka [31]

Answer: 2000

Step-by-step explanation:

Simple interest is calculated as:

(Principal × Rate × Time) / 100

We then slot the value into the formula. This would be:

720 = (P × 6 × 6)/100

720 × 100 = 36P

72000 = 36P

Principal = 72000/36

Principal = 2000

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3 years ago

If an Alaskan mountain is over 20,300 ft high , then it is the highest in Alaska .

g100num [7]

That means that there is a mountain in Alaska that is 20,300 ft high!

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3 years ago

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MrRa [10]

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2 years ago

The sensitivity is about 0.993. That is, if someone has HIV, there is a probability of 0.993 that they will test positive. • The

seraphim [82]

Answer:

(a) The probability that someone will test positive and have HIV is 0.000025.

(b) The probability that someone will test positive and not have HIV is 0.0001.

(d) The probability a person has HIV given that he/she was tested positive is 0.1986.

Step-by-step explanation:

Denote the events as follows:

<em>X</em> = a person has HIV

<em>Y</em> = a person is tested positive for HIV.

The information provided is:

$P(Y|X)=0.993\\P(Y^{c}|X^{c})=0.9999\\P(X)=0.000025$

Compute the probability of a person not having HIV as follows:

$P(X^{c})=1-P(X)=1-0.000025=0.999975$

Compute the probability of $(Y^{c}|X)$ as follows:

$P(Y^{c}|X)=1-P(Y|X)=1-0.993=0.007$

Compute the probability of as $(Y|X^{c})$ follows:

$P(Y|X^{c})=1-P(Y^{c}|X^{c})=1-0.9999=0.0001$

(a)

Compute the probability that someone will test positive and have HIV as follows:

$P(Y\cap X)=P(Y|X)P(X)\\=0.993\times0.000025\\=0.000024825\\\approx0.000025$

Thus, the probability that someone will test positive and have HIV is 0.000025.

(b)

Compute the probability that someone will test positive and not have HIV as follows:

$P(Y\cap X^{c})=P(Y|X^{c})P(X^{c})\\=0.0001\times0.999975\\=0.0000999975\\\approx0.0001$

Thus, the probability that someone will test positive and not have HIV is 0.0001.

(c)

Compute the probability that someone will test positive as follows:

$P(Y)=P(Y\cap X)+P(Y\cap X^{c})=0.000025+0.0001=0.000125$

Thus, the probability that someone will test positive is 0.000125.

(d)

Compute the probability a person has HIV given that he/she was tested positive as follows:

$P(X|Y)=\frac{P(Y|X)P(X)}{P(Y)} \\=\frac{0.993\times0.000025}{0.000125}\\ =0.1986$

Thus, the probability a person has HIV given that he/she was tested positive is 0.1986.

As the probability of a person having HIV given that he was tested positive is not very large, it would not be wise to implement a random testing policy.

3 0

4 years ago