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Vaselesa [24]
3 years ago
12

Given variables x, y, and z. What is the end value of y?

Mathematics
1 answer:
Fynjy0 [20]3 years ago
6 0

9514 1404 393

Answer:

  B. 10

Step-by-step explanation:

If this is a list of statements in a programming language that is processed sequentially, y is assigned the value 9+1 = 10. There is no further assignment to y, even after the value of x is changed. So, the final value of y is 10.

__

<em>Additional comment</em>

Some programming languages do their processing in parallel, so that the definition of y would be updated when the value of x gets updated. In that case, the value of y would be changed to 5+1 = 6. Since this is not a choice, the problem is apparently supposed to be interpreted as above.

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600 coins and 240 are pennies. What percentage are pennies?
AleksAgata [21]
240 out of 600
240 pennies out of 600 pennies = 40% 
5 0
3 years ago
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What does 8,972 ÷9 =​
Neporo4naja [7]

Answer:

It is 4486.

Step-by-step explanation:

7 0
3 years ago
We wish to give a 90% confidence interval for the mean value of a normally distributed random variable. We obtain a simple rando
coldgirl [10]

Answer: (9.27025,\ 11.12975)

Step-by-step explanation:

Given : Sample size : n= 9

Degree of freedom = df =n-1 =8

Sample mean : \overline{x}=10.2

sample standard deviation : s= 1.5

Significance level ; \alpha= 1-0.90=0.10

Since population standard deviation is not given , so we use t- test.

Using t-distribution table , we have

Critical value = t_{\alpha/2, df}=t_{0.05 , 8}=1.8595

Confidence interval for the population mean :

\overline{x}\pm t_{\alpha/2, df}\dfrac{s}{\sqrt{n}}

90% confidence interval for the mean value will be :

10.2\pm (1.8595)\dfrac{1.5}{\sqrt{9}}

10.2\pm (1.8595)\dfrac{1.5}{3}

10.2\pm (1.8595)(0.5)

10.2\pm (0.92975)

(10.2-0.92975,\ 10.2+0.92975)

(9.27025,\ 11.12975)

Hence, the 90% confidence interval for the mean value= (9.27025,\ 11.12975)

6 0
3 years ago
30 is what percent of 200?​
madam [21]

Answer:

15 percent

Step-by-step explanation:

  • 30/200=15/100
  • 15/100=0.15
  • 0.15=15 percent
5 0
3 years ago
Consider the following frequency distribution.
Zarrin [17]

Answer:

Class interval                            10-19 20-29 30-39 40-49 50-59

cumulative frequency               10      24       41        48       50

cumulative relative frequency 0.2   0.48     0.82    0.96      1

Step-by-step explanation:

1.

We are given the frequency of each class interval and we have to find the respective cumulative frequency and cumulative relative frequency.

Cumulative frequency

10

10+14=24

14+17=41

41+7=48

48+2=50

sum of frequencies is 50 so the relative frequency is f/50.

Relative frequency

10/50=0.2

14/50=0.28

17/50=0.34

7/50=0.14

2/50=0.04

Cumulative relative frequency

0.2

0.2+0.28=0.48

0.48+0.34=0.82

0.82+0.14=0.96

0.96+0.04=1

The cumulative relative frequency is calculated using relative frequency.

Relative frequency is calculated by dividing the respective frequency to the sum of frequency.

The cumulative frequency is calculated by adding the frequency of respective class to the sum of frequencies of previous classes.

The cumulative relative frequency is calculated by adding the relative frequency of respective class to the sum of relative frequencies of previous classes.

4 0
3 years ago
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