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3 years ago

4²(20-x) -8(x+3) =8 Please show your work!

1 answer:

7 0

You would do 4×4 and get 16(20-x)-8(x+3)=8 so then × 16 by 20 and 16 by x and get 320-16x and then × -8 by x and -8 by 3 and get -8x-24 so then you would have 320-16x--8x-24=8 then - -8x from -8x and -16x and end up with 320-8x-24=8 then subtract 320-24 and get 296 and then have 296-8x=8 then subtract 296 on both sides and you will get -8x=288 then divide -8 on both sides and x= 37

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what is the probability of rolling 15 with a pair of dice? answer in likely, unlikely, equally likely, or impossible

aleksley [76]

it is clearly impossible

8 0

3 years ago

A basketball gymnasium is 25 meters high, 80 meters wide and 200 meters long. We want to connect two strings, one from each of t

Rudiy27

Answer:

a) the centre is at (12.5 m, 40 m , 100 m ) with respect to our position

b) the length of the strings S will be 216.85 m

c) the angle that is formed by the strings is 1.23 rad

Step-by-step explanation:

assuming that we stand on one of the corners on the floor , so our coordinates are (0,0,0) , then the coordinates of the center of the gymnasium are found through

x center = (25 + 0)/2 = 12.5 m

y center = (80+ 0)/2 = 40 m

z center = (200+ 0)/2 = 100 m

then the centre is at (12.5 m, 40 m , 100 m ) with respect to our position

b) the length of the strings S will be the modulus of the vector that points from our position to the diagonally opposite corners

|S| = √(25²+80²+200²) = 216.85 m

c) the angle can be found through the dot product of the vectors that represent the strings S₁ and S₂

S₁ =(25,80,10)

S₂ =(-25,80,100)

then

S₁*S₂ = 25*(-25) +80*80 + 100*100 = 15775

but also

S₁*S₂ = |S₁||S₂| cos θ = |S|² * cos θ

S₁*S₂ = |S|² * cos θ

cos θ= S₁*S₂/|S|²

θ= cos ⁻¹ ( S₁*S₂/|S|² ) = cos ⁻¹ [15775/(25²+80²+200²)] = 1.23 rad

7 0

3 years ago

A. B. C. D. Whats the answer

icang [17]

I think it’s c not so sure tho

3 0

3 years ago

Read 2 more answers

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

4 years ago

How to solve this equation?

lilavasa [31]

I solved it on your other posting of it. x = 12
All of the steps are there

7 0

3 years ago

Read 2 more answers