Lanny’s credit card has an apr of 33%, calculated on the previous monthly balance. His credit card record for the last 7 months

The capacity is approximately 3.5 fluid ounces. In order to determine this, we need to compute the volume of a cone-shaped cup. The formula for the volume of a cone is: 1/3 * π * r^2 * h, where r equals the diameter divided by 2, which gives 1.35 inches, and h equals 3.3 inches. After substituting these values, we find the volume to be V = 1/3 * 3.14 * 1.35^2 * 3.3 = 6.3 cubic inches. To convert cubic inches to fluid ounces, we use the relationship that 1 fluid ounce is equal to 1.8 cubic inches. Therefore, x fluid ounces equal to 6.3 cubic inches leads to x = 6.3 / 1.8, which results in 3.5 fluid ounces.

# of guests   waitstaff gratuity
10                      18
20                      36
30                      54
40                      72
50                      90

<span>If the gratuity changes directly with the number of guests, which formula outlines the connection between the gratuity, t, and the number of guests, g? 
t = 1.8g t = g/18 g = 1.8t g = t/18
</span>
18/10 = 1.8
36/20 = 1.8
54/30 = 1.8

direct variation: <span>t = 1.8g</span>

Answer:

89

Step-by-step explanation:

Given that

2 O, 2 T and 1 M

Now based on this, the following arrangements exist  

1

Three arrangements i.e. {M,T,O}

2

XX or XY

XX in 2C1 = two arrangements i.e. {OO or TT}

XY in 3C2 × 2! = six arrangements

3

XXY or XYZ

XXY in 2C1 × 2C1 × 3! ÷ 2! = twelve arrangements

XYZ in 3C3 × 3! = six arrangements

4

XXYY or XXYZ

XXYY = 4! ÷ (2! × 2!) = six arrangements

XXYZ in 2C1 ×  4! ÷ 2! = twenty four arrangements

5

= 5! ÷ (2! ×2!)

= 120 ÷ 4

= 30

Thus, the overall total is

= 3 + (2 +6)+ (12 +6) + (6 +24) + 30

= 89

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%.

a) The area of a rectangle is calculated by multiplying the length by the width:
   A = l·w

This formula allows us to express the width in terms of area and length. By dividing A by l, we find
   A/l = w

We also recognize that the perimeter of a rectangle is the total length around it.
   P = l + w + l + w
   P = 2(l + w)
We aim to rewrite the perimeter formula to isolate l on one side, using the expression for w derived earlier.
   P = 2(l + A/l)

By substituting the known value for A, we can express p(l) as
   p(l) = 2(l + 25/l)
   p(l) = 2l + 50/l


b) For lengths exceeding widths, we have
   l > w
   l > A/l
   l² > A
   l > √A
   l > √25

This indicates that the domain of p(l) is
   l ∈ (5, ∞)..... meters