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3 years ago

For the function f(x) = -(x + 1)2 + 4, identify the vertex, domain, and range.

Mathematics

1 answer:

Studentka2010 [4]3 years ago

3 0

f ( x ) = - ( x + 1)^2 + 4

_____________________

Vertex = ( - 1 , 4 )

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Domain = R

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Range = ( - infinite , 4 ]

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Risa is sewing around the side of a square each side of the blanket is 72 inches long h many inches of ribbon will rise need

Dennis_Churaev [7]

288 ; 72 x 4 = 288.

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3 years ago

Mandy is selling pancakes and waffles at her diner. On Monday she sold a total of 300 pancakes and waffles. If she charges $3 pe

elena-s [515]

Answer:

Mandy sold 250 pancakes and 50 waffles.

Step-by-step explanation:

From the information given, you can say that the sum of pancakes and waffles is equal to 300:

x+y=300

Also, the statement indicates that she charges $3 per pancake, $1 per waffle, and made a total of $800 and you can express this as:

3x+y=800

You have the following equations:

x+y=300 (1)

3x+y=800 (2), where:

x is the number of pancakes she sold

y is the number of waffles she sold

Now, you can solve for x in (1):

x=300-y (3)

Then, you can replace (3) in (2):

3(300-y)+y=800

900-3y+y=800

900-800=2y

100=2y

y=100/2

y=50

Finally, you can replace the value of y in (3) to find x:

x=300-50

x=250

According to this, the answer is that Mandy sold 250 pancakes and 50 waffles.

5 0

2 years ago

Please, quickly. Sorry for the poor picture quality.

Nimfa-mama [501]

Answer:

<2 is 50º

Step-by-step explanation:

you first want to find what x is, so you would put the two equations in an (equation)=(equation) and get x and the number across from each other, that would be x=15º. you then plug in the 15 to the equation, and solve that, resulting in 130º. The angle is going to be 180 in total, so subtract the 130º from 180º and you would get 50º

3 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].

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