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3 years ago

7

A study shows that 78% of adults need eye correction. If 12 adults are randomly selected, find the probability that atleast 11 o

f them need correction for their eyesight

1 answer:

sergejj [24]3 years ago

7 0

Answer: 0.2224

Step-by-step explanation:

Given: The proportion of adults need eye correction: p= 78%=0.78

Let X be a binomial variable that represents the number of adults who need eye correction.

Sample size of adults: n= 12

Binomial ditsribution formula:

$P(X=x)=^nC_xp^x(1-p)^{n-x}$

Now, the probability that at least 11 of them need correction for their eyesight will be

$P(X\geq11)=P(X=11)+P(X=12)\\\\=^{12}C_{11}(0.78)^{11}(1-0.78)^{1}+^{12}C_{12}(0.78)^{12}(1-0.78)^{0}\\\\=(12)(0.78)^{11}(0.22)+(1)(0.78)^{12}(1)\\\\\approx0.1716503+0.05071486=0.22236516\approx0.2224$

Hence, the required probability = 0.2224

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A bicycle has a listed price of $643.95 before tax. If the sales tax rate is 6.75%, find the total cost of the bicycle with sale

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Answer:

$687.42

Step-by-step explanation:

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2 years ago

Kyle is 12 years older than Susan. Last year, he was twice as old as Susan.

iogann1982 [59]

Answer:

13 and 25

Step-by-step explanation:

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3 years ago

Petra jogs 3 miles in 30 minutes at this rate how long would it take a jog 10 miles

bazaltina [42]

Answer:

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Step-by-step explanation:

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7 0

3 years ago

Read 2 more answers

Solve these linear equations by Cramer's Rules Xj=det Bj / det A:

timurjin [86]

Answer:

(a) $x_1=-2,x_2=1$

(b) $x_1=\frac{3}{4} ,x_2=-\frac{1}{2} ,x_3=\frac{1}{4}$

Step-by-step explanation:

(a) For using Cramer's rule you need to find matrix $A$ and the matrix $B_j$ for each variable. The matrix $A$ is formed with the coefficients of the variables in the system. The first step is to accommodate the equations, one under the other, to get $A$ more easily.

$2x_1+5x_2=1\\x_1+4x_2=2$

$\therefore A=\left[\begin{array}{cc}2&5\\1&4\end{array}\right]$

To get $B_1$ , replace in the matrix A the 1st column with the results of the equations:

$B_1=\left[\begin{array}{cc}1&5\\2&4\end{array}\right]$

To get $B_2$ , replace in the matrix A the 2nd column with the results of the equations:

$B_2=\left[\begin{array}{cc}2&1\\1&2\end{array}\right]$

Apply the rule to solve $x_1$ :

$x_1=\frac{det\left(\begin{array}{cc}1&5\\2&4\end{array}\right)}{det\left(\begin{array}{cc}2&5\\1&4\end{array}\right)} =\frac{(1)(4)-(2)(5)}{(2)(4)-(1)(5)} =\frac{4-10}{8-5}=\frac{-6}{3}=-2\\x_1=-2$

In the case of B2, the determinant is going to be zero. Instead of using the rule, substitute the values of the variable $x_1$ in one of the equations and solve for $x_2$ :

$2x_1+5x_2=1\\2(-2)+5x_2=1\\-4+5x_2=1\\5x_2=1+4\\ 5x_2=5\\x_2=1$

(b) In this system, follow the same steps,ust remember $B_3$ is formed by replacing the 3rd column of A with the results of the equations:

$2x_1+x_2 =1\\x_1+2x_2+x_3=0\\x_2+2x_3=0$

$\therefore A=\left[\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right]$

$B_1=\left[\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right]$

$B_2=\left[\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right]$

$B_3=\left[\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right]$

$x_1=\frac{det\left(\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{1(2)(2)+(0)(1)(0)+(0)(1)(1)-(1)(1)(1)-(0)(1)(2)-(0)(2)(0)}{(2)(2)(2)+(1)(1)(0)+(0)(1)(1)-(2)(1)(1)-(1)(1)(2)-(0)(2)(0)}\\ x_1=\frac{4+0+0-1-0-0}{8+0+0-2-2-0} =\frac{3}{4} \\x_1=\frac{3}{4}$

$x_2=\frac{det\left(\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{(2)(0)(2)+(1)(0)(0)+(0)(1)(1)-(2)(0)(1)-(1)(1)(2)-(0)(0)(0)}{4} \\x_2=\frac{0+0+0-0-2-0}{4}=\frac{-2}{4}=-\frac{1}{2}\\x_2=-\frac{1}{2}$

$x_3=\frac{det\left(\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)}=\frac{(2)(2)(0)+(1)(1)(1)+(0)(1)(0)-(2)(1)(0)-(1)(1)(0)-(0)(2)(1)}{4} \\x_3=\frac{0+1+0-0-0-0}{4}=\frac{1}{4}\\x_3=\frac{1}{4}$

6 0

4 years ago

Which ratios approximate pi ()?

kramer

9514 1404 393

Answer:

A: 66/21

E: 22/7

Step-by-step explanation:

The offered choices reduce to 3 1/7 or to 3. The best approximations of those given are the ones that reduce to 3 1/7.

A: 66/21 = 22/7 = 3 1/7

B: 60/20 = 3

C: 21/7 = 3

D: 45/15 = 3

E: 22/7 = 3 1/7

___

<em>Comment on pi approximations</em>

The ratio 22/7 differs from π by about 0.04%. A better approximation is 355/113, which differs from π by about 0.000008%. The approximation 3.14 differs from π by about 0.05%.

5 0

3 years ago