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statuscvo [17]
1 year ago
8

Maya collected data about the number of ice cubes and milliliters of juice in several glasses of juice and organized the data in

to this table.
3
Ice Cubes
Juice (milliliters)
2
234
3
202
5
1 40
5
155
177
210
265
She used a graphing tool to display the data in a scatter plot, with x representing the number of ice cubes and yrepresenting the milliliters of
juice. Then she used the graphing tool to find the equation of the line of best fit:
y=-29.202x+ 293.5.
Based on the line of best fit, approximately how many milliliters of juice will be in a glass with 7 ice cubes?
A
10
B. 89
OC.
D. 208
Mathematics
1 answer:
Alex_Xolod [135]1 year ago
5 0

Answer: B. 89

Step-by-step explanation:

-29.202x7= -204.414

-204.414+293.5= 89.086

So the answer is B. 89

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Answer:

5th grade mean - 4.67

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se the function to show that fx(0, 0) and fy(0, 0) both exist, but that f is not differentiable at (0, 0). f(x, y) = 9x2y x4 + y
alexandr1967 [171]

Answer:

It is proved that f_x, f_y exixts at (0,0) but not differentiable there.

Step-by-step explanation:

Given function is,

f(x,y)=\frac{9x^2y}{x^4+y^2}; (x,y)\neq (0,0)

  • To show exixtance of f_x(0,0), f_y(0,0) we take,

f_x(0,0)=\lim_{h\to 0}\frac{f(h+0,k+0)-f(0,0)}{h}=\lim_{h\to 0}\frac{\frac{9h^2k}{h^4+k^2}-0}{h}\\\therefore f_x(0,0)=\lim_{h\to 0}\frac{9hk}{h^4+k^2}=\lim_{h\to 0}\frac{9k}{h^3+\frac{k^2}{h}}=0    exists.

And,

f_y(0,0)=\lim_{k\to 0}\frac{f(h,k)-f(0,0)}{k}=\lim_{k\to 0}\frac{9h^2k}{k(h^4+k^2)}=\lim_{k\to 0}\frac{9h^2}{h^4+k^2}=\frac{9}{h^2}   exists.

  • To show f(x,y) is not differentiable at the origin cheaking continuity at origin be such that,

\lim_{(x,y)\to (0,0)}\frac{9x^2y}{x^4+y^2}=\lim_{x\to 0\\ y=mx^2}\frac{9x^2y}{x^4+y^2}=\frac{9x^2\times m x^2}{x^4+m^2x^4}=\frac{9m}{1+m^2}  where m is a variable.

which depends on various values of m, therefore limit does not exists. So f(x,y) is not continuous at (0,0). Hence it is not differentiable at (0,0).

4 0
2 years ago
Raj wants to take yoga classes. There are two yoga studios that have different membership plans.
Vilka [71]

C

B

your welcome i took test sorry if wrong

7 0
1 year ago
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Joan is hiking. She wants to cover 1,600 feet. She hikes 600 feet every 3 hours.
Stella [2.4K]
Joan's remaining distance is reduced by (600 ft)/(3 hours) = 200 ft/hour. She starts with 1600 ft remaining, so her distance remaining (y) after x hours is
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6 0
1 year ago
A company employs two shifts of workers. Each shift produces a type of gasket where the thickness is the critical dimension. The
Oksanka [162]

Answer:

a) [-0.134,0.034]

b) We are uncertain

c) It will change significantly

Step-by-step explanation:

a) Since the variances are unknown, we use the t-test with 95% confidence interval, that is the significance level = 1-0.05 = 0.025.

Since we assume that the variances are equal, we use the pooled variance given as

s_p^2 = \frac{ (n_1 -1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2},

where n_1 = 40, n_2 = 30, s_1 = 0.16, s_2 = 0.19.

The mean difference \mu_1 - \mu_2 = 10.85 - 10.90 = -0.05.

The confidence interval is

(\mu_1 - \mu_2) \pm t_{n_1+n_2-2,\alpha/2} \sqrt{\frac{s_p^2}{n_1} + \frac{s_p^2}{n_2}} = (-0.05) \pm t_{68,0.025} \sqrt{\frac{0.03}{40} + \frac{0.03}{30}}

= -0.05\pm 1.995 \times 0.042 = -0.05 \pm 0.084 = [-0.134,0.034]

b) With 95% confidence, we can say that it is possible that the gaskets from shift 2 are, on average, wider than the gaskets from shift 1, because the mean difference extends to the negative interval or that the gaskets from shift 1 are wider, because the confidence interval extends to the positive interval.

c) Increasing the sample sizes results in a smaller margin of error, which gives us a narrower confidence interval, thus giving us a good idea of what the true mean difference is.

6 0
2 years ago
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