Jeremiah is saving money from a tutoring job.After the first three weeks,he saved $135.Assume the situation is proportional.Use
1 answer:
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Answer:
At h=0.1, the calculated value of y(1.2) equals 15.920, and for h=0.05, y(1.2) amounts to 16.523.
Step-by-step explanation:
Given,
with y(1)=9.
This means when x=1, y=9 initially.
To determine the value of y= f(x,y) at when h=0.1 and h=0.05, we will implement a C program using Euler's method resulting in, for h=0.1:
#include<stdio.h>
float fun(float x, float y)
{
float f;
f=2x+3y+1;
return f;
}
main()
{
float a,b,x,y,h,t,k;
printf("\ Enter x0,y0,h,xn:");
scanf("% f % f % f % f ", & a, &b, &h, &t);
x=a;
y=b;
printf("\n x\t y\n");
while (x<=t)
{ k=h*fun(x,y); x=x+h; y=y+k; printf("%0.3f\%0.3f\n",x,y);}}
Using initial values x0=1, y0=9, h=0.1, and xn=1.2, we reach results, y(1.2)=15.920.
Subsequently, using values x0=1, y0=9, with h=0.05, and xn=1.2 yields y(1.2)=16.523.
Utilize the details to create inequalities that reflect each limitation or requirement.
2) Label the variables.
c: count of color copies
b: count of black-and-white copies
3) Define each constraint:
i) <span>Printing a color copy requires 3 minutes while a black-and-white copy takes 1 minute.
</span><span>
</span><span>3c + b
</span><span>
</span><span>
</span><span>ii) He must print a minimum of 6 copies ⇒ c + b ≥ 6
</span><span>
</span><span>
</span><span>iv) Moreover, he must finish the prints within 12 minutes at most ⇒</span>
3c + b ≤ 12
<span />
4) Additional limits include c ≥ 0, and b ≥ 0 (meaning only non-negative counts are valid for each type of copy)
5) Here’s how to illustrate that:
i) For 3c + b ≤ 12: draw the line representing 3c + b = 12 and shade the area above and to the right of this line.
ii) For c + b ≥ 6: draw the line c + b = 6 and shade the area below and to the left of this line.
iii) Given that c ≥ 0 and b ≥ 0, the relevant region is located in the first quadrant.
iv) The concluding area is the overlap of the previously mentioned shaded regions.
v) The graph can be viewed in the attached figure.
2) Label the variables.
c: count of color copies
b: count of black-and-white copies
3) Define each constraint:
i) <span>Printing a color copy requires 3 minutes while a black-and-white copy takes 1 minute.
</span><span>
</span><span>3c + b
</span><span>
</span><span>
</span><span>ii) He must print a minimum of 6 copies ⇒ c + b ≥ 6
</span><span>
</span><span>
</span><span>iv) Moreover, he must finish the prints within 12 minutes at most ⇒</span>
3c + b ≤ 12
<span />
4) Additional limits include c ≥ 0, and b ≥ 0 (meaning only non-negative counts are valid for each type of copy)
5) Here’s how to illustrate that:
i) For 3c + b ≤ 12: draw the line representing 3c + b = 12 and shade the area above and to the right of this line.
ii) For c + b ≥ 6: draw the line c + b = 6 and shade the area below and to the left of this line.
iii) Given that c ≥ 0 and b ≥ 0, the relevant region is located in the first quadrant.
iv) The concluding area is the overlap of the previously mentioned shaded regions.
v) The graph can be viewed in the attached figure.
Answer:
Carmen's estimate is too low, since rolling 200 times results in a total of 50.
Step-by-step explanation:
Initially, we will define the sample space for this scenario.
The sample space is Ω = {1,2,3,4,5,6,7,8}
For the event A: "Rolling an 8-sided die to get a multiple of 4"
The probability for event A is
This is because there are two multiples of 4 (4 and 8) out of a total of eight numbers (1 through 8).
Next, considering the random variable X: "Total count of multiples of 4 if she rolls an 8-sided die 200 times"
X can be described as following a Binomial distribution.
X ~ Bi (n,p)
X ~ Bi (200,)
Where n is the number of rolls and p is the probability of success, defined as rolling a multiple of 4.
The mean for this variable is
Thus, we conclude that Carmen's prediction is low, as rolling 200 times yields 50.