HELP!!! ASAP!!! YOU WILL GET MARKED AS BRAINLIEST!!! You expand a rectangular garden so the perimeter is now twice the perimeter

I will designate the hourly rate for weekdays as x and for weekends as y. The equations are arranged as follows:

13x + 14y = $250.90
15x + 8y = $204.70

This gives us a system of equations which can be solved by multiplying the first equation by 4 and the second by -7. This leads to:

52x + 56y = $1003.60
-105x - 56y = -$1432.90

By summing these two equations, we arrive at:

-53x = -$429.30 --> 53x = $429.30 --> (dividing both sides by 53) x = 8.10. This represents her hourly wage on weekdays.

Substituting our value for x allows us to determine y. I will utilize the first equation, but either could work.

$105.30 + 14y = $250.90. To isolate y, subtract $105.30 from both sides --> 14y = $145.60 divide by 14 --> y = $10.40

Thus, we find that her earnings are $8.10 per hour on weekdays and $10.40 per hour on weekends. The difference shows she earns $2.30 more on weekends than on weekdays.

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%.

<span>As the restaurant owner, The likelihood of hiring Jun is 0.7 => p(J) The likelihood of hiring Deron stands at 0.4 => p(D) The chance of hiring at least one of them is 0.9 => p(J or D) We can formulate the probability equation: p(J or D) = p(J) + p(D) - p(J and D) => 0.9 = 0.7 + 0.4 - p(J and D) p(J and D) = 1.1 - 0.9 = 0.2 Thus, the probability that both Jun and Deron are hired is 0.2.</span>

I believe that option C is the appropriate choice from the provided selections. The term 'rem preimage' does not accurately characterize polygon A'B'C'D'. The preimage is defined as the original shape before undergoing any transformation. In this scenario, the transformation involved is a rotation combined with a dilation. I hope this clarifies your inquiry.