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Answer:
(x, y) = (7, -1)
Step-by-step explanation:
The expression for x makes it easy to use that for substitution.
3(-7y) +y = 20
-20y = 20 . . . . . collect terms
y = -1 . . . . . . . . . divide by -20
-7(-1) = x = 7 . . . use y in the second equation to find x
The solution is (x, y) = (7, -1).
Answer:
Lavania observed 39 fruit flies after 6 days of observation
Step-by-step explanation:
Let x be the number of fruit flies on the first day of Lavania's study.
After 6 days she had nine more than five times as many fruit flies as when she began the study.
Five times as many fruit flies as when she began the study = 5x
Nine more than five times as many fruit flies as when she began the study=5x+9
The expression to find the population of fruit flies Lavania observed after 6 days is 5x+9
If she observes 20 fruit flies on the first day of the study, then x=6, then

Answer:
The correct answer is (d) 36 units.
Step-by-step explanation:
Just took the test on edge :)
B.
You know that the angle formed by 51 and y + 6 is equal to 106 because it needs to be supplementary with the 74. Thus, after subtracting you can get 49.
Answer:
The correct option is (b).
Step-by-step explanation:
The (1 - <em>α</em>)% confidence interval for population mean (<em>μ</em>) is:

The confidence interval for population mean can be computed using either the <em>z</em>-interval or <em>t</em>-interval.
The <em>t</em>-interval is used if the following conditions are satisfied:
- The population standard deviation is not known
- The sample size is large enough
- The population from which the sample is selected is normally distributed.
For computing a (1 - <em>α</em>)% confidence interval for population mean , it is necessary for the population to normally distributed if the sample selected is small, i.e.<em>n</em> < 30, because only then the sampling distribution of sample mean will be approximated by the normal distribution.
In this case the sample size is, <em>n</em> = 28 < 30.
Also it is provided that the systolic blood pressure is known to have a skewed distribution.
Since the sample is small and the population is not normally distributed, the sampling distribution of sample mean will not be approximated by the normal distribution.
Thus, no conclusion can be drawn from the 90% confidence interval for the mean systolic blood pressure.
The correct option is (b).