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3 years ago

. Let A = {x: x ϵ R, x2 – 5x + 6 = 0 } and B = { x: x ϵ R, x2 = 9}. Find A intersection B and A union B

1 answer:

7 0

$x^2 – 5x + 6 = 0\\x^2-2x-3x+6=0\\x(x-2)-3(x-2)=0\\(x-3)(x-2)=0\\x=3 \vee x=2\\A=\{-2,3\}\\\\x^2=9\\x=-3 \vee x=3\\B=\{-3,3\}\\\\A\cap B=\{3\}\\A\cup B=\{-3,-2,3\}$

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A certain group of women has a 0.86% rate of red/green color blindness. If a woman is randomly selected, what is the probabil

Mariana [72]

Answer:

The probability that a randomly selected woman has red/green color blindness is 0.9914

Step-by-step explanation:

Given that the proportion of women having red/green color blindness is 0.86%

Represent that with p

$p= 0.86\%$

Convert proportion from percentage to decimal;

$p= \frac{0.86}{100}$

$p= 0.0086$

In probability; opposite probabilities add up to 1;

Let the probability that a woman selected have red/green color blindness be represented with q;

$p + q = 1$

Subtract p from both sides

$p - p + q = 1 - p$

$q = 1 - p$

Substitute 0.0086 for p

$q = 1 - 0.0086$

$q = 0.9914$

Hence, the probability that a randomly selected woman has red/green color blindness is 0.9914

3 0

2 years ago

The attention span of children (ages 3 to 5) is claimed to be Normally distributed with a mean of 15 minutes and a standard devi

Agata [3.3K]

Answer:

If the observed sample mean is greater than 17.07 minutes, then we would reject the null hypothesis.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 15 minutes

Sample size, n = 10

Alpha, α = 0.05

Population standard deviation, σ = 4 minutes

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 15\text{ minutes}\\H_A: \mu > 15\text{ minutes}$

Since the population standard deviation is given, we use one-tailed z test to perform this hypothesis.

Formula:

$z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }$

Now, $z_{critical} \text{ at 0.05 level of significance for one tail} = 1.64$

Thus, we would reject the null hypothesis if the z-statistic is greater than this critical value.

Thus, we can write:

$z_{stat} = \displaystyle\frac{\bar{x} - 15}{\frac{4}{\sqrt{10}} } > 1.64\\\\\bar{x} - 15>1.64\times \frac{4}{\sqrt{10}}\\\bar{x} - 15>2.07\\\bar{x} > 15+2.07\\\bar{x} > 17.07$

Thus, the decision rule would be if the observed sample mean is greater than 17.07 minutes, then we would reject the null hypothesis.

6 0

3 years ago

PLEASE HELP!!! a 300 cm b 60 cm c 20 cm d 45 cm

brilliants [131]

Answer:

45 cm

Step-by-step explanation:

Let length of the field in the drawing = x

80 : 120 :: 30 : x

Product of extreme = Product of means

80 * x = 120 * 30

$x = \frac{120*30}{80}\\\\x = 15 * 3\\x = 45$

3 0

2 years ago

The circle below has center P, and its radius is 6in. Given that =m∠QPR170°, find the length of the major arc QSR.

kenny6666 [7]

The length of the major arc QSR is 39.77 inches

<h3>How to find the length of the major arc QSR?</h3>

The given parameters are:

m∠QPR = 170 degree

Radius, r = 6 inches

The length of the major arc QSR is calculated as:

Arc length = (360 - Angle/360) * 2πr

Substitute the known values in the above equation

Arc length = (360 - 170/360) * 2 * 3.14 * 6

Evaluate the difference

Arc length = (190/180) * 2 * 3.14 * 6

Evaluate the product

Arc length = 39.77

Hence, the length of the major arc QSR is 39.77 inches