Which of these figures has rotational symmetry?

Answer:

0.2 = 20%

Step-by-step explanation:

The combinations that yield a total of 6 or lower include:

(1,1), (1,2), (1,3), (1,4), (1,5),

(2,1), (2,2), (2,3), (2,4),

(3,1), (3,2), (3,3),

(4,1), (4,2),

(5,1)

(15 combinations in total)

The combinations resulting in a total of 4 are:

(1,3), (2,2), (3,1)

(3 combinations in total)

Thus, the probability of obtaining a sum of 4, given that the total is 6 or less is:

P = 3/15 = 0.2 = 20%

Response:

a. 55 cars

b. 25 cars

Detailed explanation:

Let’s denote the quantity of cars with stereo systems as N(ss), those with air conditioning as N(ac), and those with sunroofs as N(sr).

We find that:

N(ss) = 30

N(ac) = 30

N(sr) = 40

N(ss and ac and sr) = 15

N(at least two) = 30

a.

To calculate how many cars possess at least one feature (N(at least one) or N(ss or ac or sr)), we apply:

N(ss or ac or sr) = N(ss) + N(ac) + N(sr) - N(ss and ac) - N(ss and sr) - N(ac and sr) + N(ss and ac and sr)

N(ss or ac or sr) = 30 + 30 + 40 - (N(at least two) + 2*N(ss and ac and sr)) + 15

Substituting, we find N(ss or ac or sr) = 30 + 30 + 40 - (30 + 2*15) + 15 = 55

b.

For those cars that have exactly one feature, we have:

N(only one) = N(at least one) - N(at least two)

N(only one) = 55 - 30 = 25

The required work to pump water is 3,325,140 Joules. Step-by-step explanation: Work done by the pump is calculated by multiplying the force exerted on the pump by the distance the water is moved. Force equates to mass multiplied by gravitational acceleration. Consequently, Force = (water density × tank volume) × gravitational acceleration, leading to F = ρVg. Therefore, Work done = (ρVg) * d. Given the values of ρ = 1000 kg/m³, g = 9.8 m/s², d = 3 m, we compute the work done: Work = 1000 * 113.10 * 9.8 * 3 = 3,325,140 Joules.

Response:

y=-10x

Detailed explanation:

y=-10x

y/x=-10

Answer: (97.98, 112.020)

Step-by-step explanation: We will create a 95% confidence interval for the average weight of melons.

Given the information, we determine that the critical value for the interval needs to be retrieved from a t distribution table due to the sample size being below 30 (specifically, 20), and we are provided with the sample standard deviation (s = 15 lb).

The parameters provided are:

Sample mean = x = 105 lb

Sample standard deviation = s = 15 lb

Sample size = n = 20

To establish the 95% confidence interval, we indicate that the level of significance is 5%.

The formula for the confidence interval is:

u = x + tα/2 × s/√n... for the upper limit

u = x - tα/2 × s/√n... for the lower limit.

tα/2 represents the critical value for the test (which will be determined using the t distribution table).

To derive tα/2, we look for the value based on the degrees of freedom (sample size - 1) against the significance level for a two-tailed test (α/2 = 0.025%) in a t distribution table.

For the upper limit, we calculate:

u = 105 + 2.093×15/√20

u = 105 + 2.093× (3.3541)

u = 105 + 7.020

u = 112.020.

<pfor the="" lower="" limit="" we="" find:="">

u = 105 - 2.093×15/√20

u = 105 - 2.093× (3.3541)

u = 105 - 7.020

u = 97.98

Confidence interval (97.98, 112.020)

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