All categories

3 years ago

Can someone simplify

Mathematics

1 answer:

madam [21]3 years ago

8 0

APP IT will HELP SIDJIDNUDNUD

You might be interested in

Using distributive property rewrite the expression without parentheses.

timofeeve [1]

6y + 30

4a - 24

6 + 2w

7 0

3 years ago

PLS ANSWER THE BELOW QUESTION AND I'LL MARK U AS BRAINLIST

MA_775_DIABLO [31]

Answer:

Step-by-step explanation:

$a) \dfrac{3x}{2}-5=4$

Add 5 to both sides

$\dfrac{3x}{2}-5 +5 = 4 + 5\\\\\dfrac{3x}{2} = 9 \\\\Now \ multiply \ both \ sides \ by \ 2 \\\\\dfrac{3x}{2}*2 = 9*2\\\\3x = 18\\Now \ divide \ both \ sides \ by \ 3\\\\x =\dfrac{18}{3}\\\\$

x = 6

$b) \dfrac{x-3}{5} + 1 = -2\\\\\\Subtract \ 1 \ from \ both \ sides\\\\\dfrac{x-3}{5} = -2 -1\\\\\dfrac{x-3}{5} = -3\\\\Multiply \ both \ side \ by \ 5\\\\x - 3 = -3*5\\\\x -3 = -15\\\\$

Add 3 to both sides

x = -15 +3

x = -12

7 0

2 years ago

Read 2 more answers

How do you find the prime factorization of 504

arlik [135]

1 x 2 x 2 x 2 x 3 x 3 x 7

6 0

3 years ago

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

How many meters are in a millimeter

Naddika [18.5K]

There are 1000 millimeters in a meter and 0.001 meters in a millimeter

7 0

3 years ago

Read 2 more answers