2/3p-1/4p when p=2.4 (I messed up last time)

A circle with radius of \greenD{5\,\text{cm}}5cmstart color #1fab54, 5, start text, c, m, end text, end color #1fab54 sits insid

Marat540 [252]

Answer:

The area of the shaded region is 42.50 cm².

Step-by-step explanation:

Consider the figure below.

The radius of the circle is, r = 5 cm.

The sides of the rectangle are:

l = 11 cm

b = 11 cm.

Compute the area of the shaded region as follows:

Area of the shaded region = Area of rectangle - Area of circle

$=[l\times b]-[\pi r^{2}]\\\\=[11\times 11]-[3.14\times 5\times 5]\\\\=121-78.50\\\\=42.50$

Thus, the area of the shaded region is 42.50 cm².

7 0

2 years ago

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A certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder. In

lord [1]

Answer:

95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

P.Q. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of males having blood disorder= $\frac{250}{1000}$ = 0.25

$\hat p_2$ = sample proportion of females having blood disorder = $\frac{275}{1000}$ = 0.275

$n_1$ = sample of males = 1000

$n_2$ = sample of females = 1000

$p_1$ = population proportion of males having blood disorder

$p_2$ = population proportion of females having blood disorder

Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.

So, 95% confidence interval for the difference between the population proportions, ( $p_1-p_2$ ) is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ${(\hat p_1-\hat p_2)-(p_1-p_2)}$ < $1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

P( $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ( $p_1-p_2$ ) < $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

95% confidence interval for ( $p_1-p_2$ ) =

[ $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ , $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ]

= [ $(0.25-0.275)-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ , $(0.25-0.275)+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ ]

= [-0.064 , 0.014]

Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

8 0

2 years ago

What value of x will make △ONM similar to △SRQ by the SAS similarity theorem?

Lerok [7]

Answer:

c 25

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Help pls and explain thank you :)

Solnce55 [7]

25 because Becky read 45 and Jason read 20

45-20 = 25

5 0

2 years ago

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What is 3/2=9/10k im having troble lots rly

tigry1 [53]

Answer:

k= 5 over 3 =1.667

Step-by-step explanation:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

(3)/(2)-((9)/(10)*k)=0

simplify 9/10

3 9

— - (—— • k) = 0

2 10

simplify 3/2

3 9k

— - —— = 0

2 10

Find the Least Common Multiple

The left denominator is : 2

The right denominator is : 10

Least Common Multiple:

10

Calculate multipliers for the two fractions

Denote the Least Common Multiple by L.C.M

Denote the Left Multiplier by Left_M

Denote the Right Multiplier by Right_M

Denote the Left Deniminator by L_Deno

Denote the Right Multiplier by R_Deno

Left_M = L.C.M / L_Deno = 5

Right_M = L.C.M / R_Deno = 1

Rewrite the two fractions into equivalent fractions

Two fractions are called equivalent if they have the same numeric value.

For example : 1/2 and 2/4 are equivalent, y/(y+1)2 and (y2+y)/(y+1)3 are equivalent as well.

To calculate equivalent fraction , multiply the Numerator of each fraction, by its respective Multiplier.

L. Mult. • L. Num. 3 • 5

—————————————————— = —————

L.C.M 10

R. Mult. • R. Num. 9k

—————————————————— = ——

L.C.M 10

Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

3 • 5 - (9k) 15 - 9k

———————————— = ———————

10 10

Pull out like factors :

15 - 9k = -3 • (3k - 5)

-3 • (3k - 5)

————————————— = 0

10

When a fraction equals zero ...

Where a fraction equals zero, its numerator, the part which is above the fraction line, must equal zero.

Now,to get rid of the denominator, Tiger multiplys both sides of the equation by the denominator.

Here's how:

-3•(3k-5)

————————— • 10 = 0 • 10

10

Now, on the left hand side, the 10 cancels out the denominator, while, on the right hand side, zero times anything is still zero.

The equation now takes the shape :

-3 • (3k-5) = 0

Solve : -3 = 0

This equation has no solution.

A a non-zero constant never equals zero.

Solve : 3k-5 = 0

Add 5 to both sides of the equation :

3k = 5

Divide both sides of the equation by 3:

k = 5/3 = 1.667

one solution was found

k = 5/3 = 1.667

hope it helps sry its so long

7 0

2 years ago

2/3p-1/4p when p=2.4 (I messed up last time)​

2/3p-1/4p when p=2.4 (I messed up last time)