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3 years ago

Simplify (√3 + 2)² OA. 7 OB. 483 OC. 4 +7√3 OD. 7+4v3 O E. 1173

Mathematics

1 answer:

fiasKO [112]3 years ago

3 0

Answer: C

Step-by-step explanation:

Simplify the radical by breaking the radicand up into a product of known factors, assuming positive real numbers.

Exact Form:7+4√3

Decimal Form:

13.92820323

Hope This Helps!

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Lily is cutting a piece of yarn into 3 (three) pieces. The 2nd piece is 3 times as long as the 1st piece, while the 3rd piece is

STatiana [176]

Answer:

The length of the first piece = 41 cm

Step-by-step explanation:

Let the length of the first piece = a

Let the length of the second piece = b

Let the length of the third piece = c

we are given the following:

b = 3a . . . . . (1) (The 2nd piece is 3 times as long as the 1st piece)

c = 6 + a . . . . (2) (the 3rd piece is 6 centimeters longer than the 1st piece)

a + b + c = 211 . . . . . (3) ( the yarn has a total length of 211 centimeters)

Next, let us eliminate two variables, and this can easily be done by substituting the values of b and c in equations 1 and 2 into equation 3. this is done as follows:

a + b + c = 211

a + (3a) + (6 + a) = 211 ( remember that b = 3a; c = 6 + a)

a + 3a + 6 + a = 211

5a + 6 = 211

5a = 211 - 6 = 205

5a = 205

∴ a = 205 ÷ 5 = 41 cm

a = 41 cm

Therefore the length of the first piece (a) = 41 cm

now finding b and c

substituting a into equation 1 and 2

b = 3a

b = 3 × 41 = 123

∴ b = 123 cm

c = 6 + a

c = 6 + 41 = 47

∴ c = 47 cm

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3 years ago

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nikdorinn [45]

Answer:

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Use the coordinates to find the length of each side of the rectangle. Then find the perimeter. L(3,3), M(3,5), N(7,5), P(7,3)

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If f(n) = -6n + 2, find f (5/6)

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$f( \frac{5}{6} ) = - 6( \frac{5}{6} ) + 2 \\ f( \frac{5}{6} ) = - \frac{30}{6} + 2 \\ f( \frac{5}{6} ) = - 5 + 2 \\ f( \frac{5}{6} ) = - 3$

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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

3 years ago