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Andrews [41]
3 years ago
15

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y = 7 − x 2 y=7-x2. what are the dime

nsions of such a rectangle with the greatest possible area?

Mathematics
1 answer:
slega [8]3 years ago
5 0
A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y = 7 − x 2 y=7-x2. what are the dimensions of such a rectangle with the greatest possible area?

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Your blood circulates a total of 128 times a hour, or 3 times a minute.
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@FauxGeoDuck
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Which two numbers are 4 units away from 0 on the number line?
Oduvanchick [21]
Answer: C


Step by step: .........
4 0
3 years ago
At an ocean-side nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water t
grandymaker [24]

Answer:

(a1) The probability that temperature increase will be less than 20°C is 0.667.

(a2) The probability that temperature increase will be between 20°C and 22°C is 0.133.

(b) The probability that at any point of time the temperature increase is potentially dangerous is 0.467.

(c) The expected value of the temperature increase is 17.5°C.

Step-by-step explanation:

Let <em>X</em> = temperature increase.

The random variable <em>X</em> follows a continuous Uniform distribution, distributed over the range [10°C, 25°C].

The probability density function of <em>X</em> is:

f(X)=\left \{ {{\frac{1}{25-10}=\frac{1}{15};\ x\in [10, 25]} \atop {0;\ otherwise}} \right.

(a1)

Compute the probability that temperature increase will be less than 20°C as follows:

P(X

Thus, the probability that temperature increase will be less than 20°C is 0.667.

(a2)

Compute the probability that temperature increase will be between 20°C and 22°C as follows:

P(20

Thus, the probability that temperature increase will be between 20°C and 22°C is 0.133.

(b)

Compute the probability that at any point of time the temperature increase is potentially dangerous as follows:

P(X>18)=\int\limits^{25}_{18}{\frac{1}{15}}\, dx\\=\frac{1}{15}\int\limits^{25}_{18}{dx}\,\\=\frac{1}{15}[x]^{25}_{18}=\frac{1}{15}[25-18]=\frac{7}{15}\\=0.467

Thus, the probability that at any point of time the temperature increase is potentially dangerous is 0.467.

(c)

Compute the expected value of the uniform random variable <em>X</em> as follows:

E(X)=\frac{1}{2}[10+25]=\frac{35}{2}=17.5

Thus, the expected value of the temperature increase is 17.5°C.

7 0
3 years ago
Can someone please help me it would mean a lot thank you.
Lubov Fominskaja [6]

Answer:

1: 0

2: 2

3: 8

4: 0

5: -19

6: 6

7: -9

8: -4

9: 10

10: 1

11: -9

12: 20

13: -29

14: 17

15: 29

16: 29

Step-by-step explanation:

They are all in order.

6 0
3 years ago
Read 2 more answers
Pschological tests are often used to determine the hostility levels in people. High scores on the HLT pschological test correspo
algol13

Answer:  the correct answer is we are 95% confident that there is no statistically significant difference in the mean treatment  methods for hostility.  

Step-by-step explanation:

1)  Test for the equality of variances in the two groups to choose the appropriate t-test.  

H0: σ (1)^2 = σ (2)^2  

Ha: σ (1)^2 ≠ σ (2)^2  

Larger variance = 64  

Smaller variance = 49  

F = 1.30612  

Degrees of freedom 15 and 9  

Critical F from the table (with alpha=0.05) = 2.58  

Calculated F is smaller than critical F, so we use the pooled variance t-test.  

Sample 1 size 7  

Sample 2 size 8  

Sample 1 mean 79  

Sample 2 mean 84  

Sample 1 S.D. 7  

Sample 2 S.D. 8  

Pooled S.D = [(n1-1)s1^2+(n2-1)s2^2]/(n1+n2-2)

Pooled variance s = [(6)(49)+(7)(64))] / (13) =  (294+448)/13=742/13=57.076923

Pooled variance s^2 = 57.076923  

Standard error of difference in means = sqrt(1/n1+1/n2) times sqrt(s^2)  

Standard error of difference in means = (0.517549)(7.554927) = 3.910046 (denominator of t)  

Confidence interval = (mean1-mean2) +/- t SE  

t is the critical t with 24 degrees of freedom = 2.056  

(79 - 84) +/- (2.056) (3.910046)  

= (-13.04, 3.04)  

Interval encloses 0  

We are 95% confident that there is no statistically significant difference in the mean treatment  methods for hostility.  

4 0
3 years ago
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