The equation of a linear function in point-slope form is y – y1 = m(x – x1). Harold correctly wrote the equation y = 3(x – 7) us
2 answers:
We need to identify the point Harold utilized for this equation:
Recall that the point-slope form of a line is expressed as:
Comparing Harold's equation to the standard form reveals the line's slope is .
Furthermore, it is clear that and
.
Therefore, Harold's equation was based on the point (7,0).
Answer:
The point (7,0) was used by Harold when writing his equation.
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This problem can be addressed by applying the normal approximation to a binomial distribution.
Calculations:
Mean (μ) = np = 10,000 × 0.5 = 5,000
The standard deviation (σ) is given by:
The probability of obtaining more than 5,100 tails is 0.0228, whereas the probability of fewer than 5,100 tails is 0.9772.
Thus, the odds of having more than 5,100 tails are:
0.0228 : 0.9772 = 1 : 42.86.
Calculations:
Mean (μ) = np = 10,000 × 0.5 = 5,000
The standard deviation (σ) is given by:
The probability of obtaining more than 5,100 tails is 0.0228, whereas the probability of fewer than 5,100 tails is 0.9772.
Thus, the odds of having more than 5,100 tails are:
0.0228 : 0.9772 = 1 : 42.86.
Answer:
We conclude that less than or equal to 50% of the student body supports him significantly.
Step-by-step explanation:
Mike, while campaigning for the student government presidency, is keen to determine if more than 50% of the student body supports him.
A random sample of 100 students was surveyed, with 55 expressing support for Mike.
Let p = the proportion of students backing Mike.
Thus, Null Hypothesis, : p
50% {indicating that the proportion of supporters among the student body is significantly less than or equal to 50%}
Alternate Hypothesis, : p > 50% {indicating that the proportion of supporters among the student body is significantly more than 50%}
The test statistics to be utilized here One-sample z proportion statistics;
T.S. = ~ N(0,1)
where, = the sampled proportion in favor of Mike =
= 0.55
n = number of sampled students = 100
Therefore, test statistics =
= 1.01
The z test statistic is 1.01.
At a significance level of 0.05, the z table provides a critical value of 1.645 for a right-tailed test.
Given that our test statistic falls below the critical value of z (1.01 < 1.645), we lack sufficient evidence to reject the null hypothesis as it remains outside the rejection region, leading to failure to reject our null hypothesis.
As a result, we conclude that less than or equal to 50% of the student body is in favor of him or the proportion supporting Mike is not significantly greater than 50%.