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hjlf
2 years ago
6

Rate at which risk of down syndrome is changing is approximated by function r(x) = 0.004641x2 − 0.3012x + 4.9 (20 ≤ x ≤ 45) wher

e r(x) measured in percentage of births per year and x is maternal age at delivery. find function f giving risk as percentage of births when maternal age at delivery is x years, given that risk of down syndrome at 30 is 0.14% of births.

Mathematics
1 answer:
romanna [79]2 years ago
7 0
The rate of change of the risk of down syndrome (in percentage of births per year) is
r(x) = 0.004641x² - 0.3012x + 4.9,   20≤ x ≤ 45
where
x = maternal age at delivery.

The function giving risk as a percentage of births when maternal age is x is the integral of r(x). That is,
f(x) = 0.001547x³ - 0.1506x² +4.9x + c

When x = 30, f = 0.14%. Therefore
0.001547(30³) - 0.1506(30²) + 4.9(30) + c = 0.14 
41.769 - 135.54 + 147 + c = 0.14
c = -53.089

Answer:
f(x) = 0.001547x³ - 0.1506x² + 4.9x - 53.089,   20 ≤ x ≤ 45

The function is graphed as shown below.

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Crazy boy [7]

Answer:

D. 13*3^x

Step-by-step explanation:

3^x +4*3^x^+^1= \\3^x+4*3(3^x)=\\3^x(1+[4*3])=\\3^x(1+12)=\\3^x(13)=\\13*3^x

3 0
2 years ago
To mix weed killer with water correctly, it is necessary to mix 18 teaspoons of weed killer with 3 gallons of water. How many ga
Alina [70]

Answer:

6 gallons of water

Step-by-step explanation:

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5 0
2 years ago
Two brands of AAA batteries are tested in order to compare their voltage. The data summary can be found below. Find the 93% conf
kobusy [5.1K]

Answer:

(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}

And replacing we got:

(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903

(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097

And the confidence interval for the difference of means would be given by:

0.2903 \leq \mu_1 -\mu_2 \leq 0.5097

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

We have the following data given:

\bar X_1 = 9.2 , \bar X_2 = 8.8, \sigma_1= 0.3, n_1 = 27, \sigma_2 = 0.1, n_2 = 30

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 93% of confidence, our significance level would be given by \alpha=1-0.93=0.07 and \alpha/2 =0.035. And the critical value would be given by:  

z_{\alpha/2}=-1.811, z_{1-\alpha/2}=1.811  

The confidence interval is given by:

(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}

And replacing we got:

(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903

(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097

And the confidence interval for the difference of means would be given by:

0.2903 \leq \mu_1 -\mu_2 \leq 0.5097

4 0
2 years ago
Use a table of numerical values of f(x,y) for (x,y) near the origin to make a conjecture about the value of the limit of f(x,y)
grigory [225]
Seems to be that the limit to compute is

\displaystyle\lim_{(x,y)\to(0,0)}\frac{xy}{x^2+2y^2}

Consider an arbitrary line through the origin y=mx, so that we rewrite the above as

\displaystyle\lim_{x\to0}\frac{mx^2}{x^2+2m^2x^2}=\lim_{x\to0}\frac m{1+2m^2}=\frac m{1+2m^2}

The value of the limit then depends on the slope m of the line chosen, which means the limit is path-dependent and thus does not exist.
8 0
2 years ago
Read 2 more answers
A 5 hour conference booking is required fir an office party for 120 people which includes a buffet dinner hotel 5 has recently i
ankoles [38]

Answer:

Total cost = $4,873

Therefore, the booking cost of hotel 5 would be $4,873

Step-by-step explanation:

Please refer to the attached table.  

The total cost includes the cost of the room and the cost of buffet dinner.

From the given table hotel 5,

The cost of the room for 120 people is found to be $166 per hour.

The conference will last for 5 hours so the total cost of the room is

Cost of room = 5*$166 = $830

From the given table hotel 5,

The cost of the buffet dinner per head is found to be $30.

Since there are total 120 people so the total cost of dinner is

Cost of dinner = 120*$30 = $3600

Total cost = $830 + $3600 = $4430

We are given that hotel 5 has recently increased the total chargers by 10%

Total cost = $4430*1.10

Total cost = $4,873

Therefore, the booking cost of hotel 5 would be $4,873.

7 0
2 years ago
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