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3 years ago

13

What output will be displayed as a result of executing the following code? int x = 5, y = 20; x += 32; y /= 4; System.out.printl

n("x = " + x + ", y = " + y);

1 answer:

Debora [2.8K]3 years ago

3 0

Answer: x = 37,y=5

Step-by-step explanation:

from the initial x is 5 and y is 20

but now the values are updated:

x+=32

x=x+32

x=5+32

x=37

so now x is 37

y/=4

y=y/4

y=20/4

y=5

so now y is 5

the out put will be:

x = 37,y=5

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How do you do this math

Basile [38]

Answer: y = -3x - 6

Step-by-step explanation:

One way to write a <u>linear equation</u> is with slope-intercept form. Slope-intercept form is y = mx + b, where m is the slope, and b is the y-intercept.

Thus, the equation is y = -3x - 6

Hope it helps :) and let me know if you are confused anywhere.

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3 years ago

Read 2 more answers

Determine the astm grain size number if 33 grains per square inch are measured at a magnification of 270×

strojnjashka [21]

8.9

The equation for the grain size is expressed as the equality:
Nm(M/100)^2 = 2^(n-1)
where
Nm = number of grains per square inch at magnification M.
M = Magnification
n = ASTM grain size number

Let's solve for n, then substitute the known values and calculate.
Nm(M/100)^2 = 2^(n-1)

log(Nm(M/100)^2) = log(2^(n-1))
log(Nm) + 2*log(M/100) = (n-1) * log(2)
(log(Nm) + 2*log(M/100))/log(2) = n-1
(log(Nm) + 2*log(M/100))/log(2) + 1 = n

(log(33) + 2*log(270/100))/log(2) + 1 = n
(1.51851394 + 2*0.431363764)/0.301029996 + 1 = n
(1.51851394 + 0.862727528)/0.301029996 + 1 = n
2.381241468/0.301029996 + 1 = n
7.910312934 + 1 = n
8.910312934 = n

So the ASTM grain size number is 8.9

If you want to calculate the number of grains per square inch, you'd use the
same formula with M equal to 1. So:
Nm(M/100)^2 = 2^(n-1)
Nm(1/100)^2 = 2^(8.9-1)
Nm(1/10000) = 2^7.9
Nm(1/10000) = 238.8564458
Nm = 2388564.458

Or about 2,400,000 grains per square inch.

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3 years ago

Find the slope of this line

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Answer:

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(Y2-Y1/X2-X1).

The answer will always be zero.

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1 year ago

Camp Canine is a kennel for boarding dogs while their owners are on vacation. The graph shows three tiers of boarding pricing. I

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Answer:

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3 years ago

A survey found that the average daily cost to rent a car in Los Angeles is $102.24 and in Las Vegas is $97.35. The data were col

cluponka [151]

Answer:

Step-by-step explanation:

The formula for determining the confidence interval for the difference of two population means is expressed as

Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)

Where

x1 = average daily cost to rent a car in Los Angeles

x2 = average daily cost to rent a car in Las Vegas

s1 = sample standard deviation for Los Angeles

s2 = sample standard deviation for Las Vegas

n1 = number of sampled cars in Los Angeles

n2 = number of sampled cars in Las Vegas

Degree of freedom = (n1 - ) + (n2 - 1) = (40 - 1) + (40 - 1) = 38

For a 95% confidence interval, the t score from the t distribution table is 2.024

From the information given,

x1 = 102.24

s1 = 5.98

n1 = 40

x2 = 97.35

s2 = 4.21

n2 = 40

x1 - x2 = 102.24 - 97.35 = 4.89

Margin of error = z√(s1²/n1 + s2²/n2) = 2.024√(5.98²/40 + 4.21²/40) = 2.024√1.3371125

= 2.34

The 95% confidence interval is 4.89 ± 2.34

Hypothesis testing

This is a test of 2 independent groups. The population standard deviations are not known. Let μ1 be the mean average daily cost to rent a car in Los Angeles and μ2 be the the mean average daily cost to rent a car in Las Vegas

The random variable is μ1 - μ2 = difference in the mean average daily cost to rent a car in Los Angeles and the mean average daily cost to rent a car in Las Vegas

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 > μ2 H1 : μ1 - μ2 > 0

This is a two tailed test

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(x1 - x2)/√(s1²/n1 + s2²/n2)

t = (102.24 - 97.35)/√(5.98²/40 + 4.21²/40)

t = 4.23

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [5.98²/40 + 4.21²/40]²/[(1/40 - 1)(5.98²/40)² + (1/40 - 1)(4.21²/40)²] = 1.78786983766/0.02552804373

df = 70

We would determine the probability value from the t test calculator. It becomes

p value = 0.00007

Since alpha, 0.05 > than the p value, 0.00007, then we would reject the null hypothesis. Therefore, at 5% significance level, there is sufficient evidence to conclude that there is a significant difference in the rates between the two cities.

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3 years ago