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The dot plots below show the test scores of sixth- and seventh-grade students: Dot plot for Grade 6 shows 6 dots on score 70, 4 dots on score 80, 6 dots on score 90, and 4 dots on score 100. Dot plot for Grade 7 shows 5 dots on score 50, 7 dots on score 60, 4 dots on score 70, and 4 dots on score 80. Based on visual inspection of the dot plots, which grade, if any, appears to have the higher mean score? Grade 6 Grade 7 Both groups show about the same mean score. No conclusion about mean score can be made from the data.

Answer:

Using theorem

AE=8.66

Thus radius of large circle is one third of equilateral triangle altitude.

Radius of larger circle=2.9 inch

And radius of inner circle will be 0.96 inch

The area of the shaded region is 42.14 square inches.

How the area of the shaded region is determined:

We are given a square with side lengths of 14 inches, and each corner of the square has a quarter circle with a radius of 7 inches. We need to find the area of the shaded region, which is the area of the square minus the area of the quarter circles.

To calculate the area of the figure, we need to consider both the square and the quarter circles.

1. Calculate the area of the square:

The square has a side length of 14 inches.

Area of the square = [tex]side^2[/tex] = [tex]14^2[/tex] = 196 square inches

2. Calculate the area of the quarter circles:

The quarter circle has a radius of 7 inches.

The area of a full circle is given by:

Area of the circle = [tex]\pi r^2[/tex]

Area of the quarter circle = 4/4 x [tex]\pi r^2[/tex]

= 1 x 3.14 x [tex]7^2[/tex]

= 153.86 square inches

Therefore, the area of the shaded region = Area of the square - Area of the quarter circles:

= 196 - 153.86

= 42.14 square inches

Complete Question:

The square in the figure has a side length of 14 inches. The radius of the each quarter circle is 7 inches. What is the area of the shaded region?

Answer:

[tex]ln(x - 1) = ln(\frac{6}{x})[/tex]

Step-by-step explanation:

[tex]ln(x - 1) = ln6 - lnx[/tex]

To solve for x we need to simplify the ln

To simplify logarithmic function we use log property

[tex]ln(a) - ln(b) = ln(\frac{a}{b})[/tex]

we apply the same property on the right hand side of the given equation

[tex]ln(x - 1) = ln6 - lnx[/tex]

[tex]ln(6) - ln(x) = ln(\frac{6}{x})[/tex]

[tex]ln(x - 1) = ln(\frac{6}{x})[/tex]

This is the first step in solving the given equation

Answer:

77.04

Step-by-step explanation:

Answer:

A) Reflect the graph of the first function across the x-axis, translate it pi/4 units to the left, and translate it 2 units up

The 90% confidence interval for the population mean [tex]\( \mu \)[/tex] is approximately (2.50, 3.21) .

To construct a 90% confidence interval for the population mean [tex]\( \mu \)[/tex], we can use the formula:

[tex]\[ \text{Confidence Interval} = \bar{x} \pm \left( \text{Critical Value} \times \frac{s}{\sqrt{n}} \right) \][/tex]

where:

- [tex]\( \bar{x} \)[/tex] is the sample mean,

- s is the sample standard deviation,

- n is the sample size, and

- the critical value corresponds to the desired confidence level and degrees of freedom.

Given:

- Sample mean [tex]\( \bar{x} = 2.86 \),[/tex]

- Sample standard deviation s=0.78

- Sample size n=15

- Confidence level = 90%.

First, we need to find the critical value corresponding to a 90% confidence level and 14 degrees of freedom (since ( n - 1 = 15 - 1 = 14 )). We can find this value using a t-distribution table or a statistical calculator. For a 90% confidence level and 14 degrees of freedom, the critical value is approximately 1.7613.

Now, let's calculate the confidence interval:

[tex]\[ \text{Confidence Interval} = 2.86 \pm \left( 1.7613 \times \frac{0.78}{\sqrt{15}} \right) \]\[ \text{Confidence Interval} = 2.86 \pm \left( 1.7613 \times \frac{0.78}{\sqrt{15}} \right) \]\[ \text{Confidence Interval} = 2.86 \pm \left( 1.7613 \times \frac{0.78}{3.87298} \right) \]\[ \text{Confidence Interval} = 2.86 \pm \left( 1.7613 \times 0.20172 \right) \]\[ \text{Confidence Interval} = 2.86 \pm 0.35587 \][/tex]

Lower Limit:

[tex]\[ 2.86 - 0.35587 \approx 2.5041 \][/tex]

Upper Limit:

[tex]\[ 2.86 + 0.35587 \approx 3.2141 \][/tex]

A mathematical verbal expression is a translation into words of an algebraic expression that can consist of different operations, numbers and varibles ur welcome.

For example, the following is a verbal phrase: "John went to the store and bought five apples. Each apple cost $1.25. How much did John spend before tax?" To translate it to a verbal model, the problem solver represents the problem in simplified words that represent the cost of the apple times the number of apples bought. To translate it further to a mathematical phrase, the problem solver puts numbers in the problem: "$1.25 times 5."

She had 2/3 quart. She used 1/4 of it.

That means you have to find 1/4 of 2/3

1/4 × 2/3 = 2/12

2/12 simplified = 1/6

Your answer is 1/6

Hope this helped!

Answer:

[tex]A=50(0.6)^x[/tex]

18 mg of medicine will be left in the patient's system after two hours.

Step-by-step explanation:

Given,

The initial quantity of the medicine, P = 50 mg,

Also, it decreases every hour at a constant rate of 40%

That is, r = 40 %,

Thus, the quantity of the medicine after x hours,

[tex]A=P(1-\frac{r}{100})^r[/tex]

[tex]=50(1-\frac{40}{100})^x[/tex]

[tex]=50(1-0.4)^x[/tex]

[tex]=50(0.6)^x[/tex]

Which is the required exponential decay function that models this scenario.

The quantity of the medicine after 2 hours,

[tex]A=50(0.6)^2=18\text{ mg}[/tex]

Answer:

195 miles

Step-by-step explanation:

50*x=125

x=2.5

2.5*78=195miles

ANSWER

[tex]{ \sec}(e) = \pm \: \sqrt{\frac{11}{8} } [/tex]

EXPLANATION

We use the Pythagorean Identity,

[tex] { \sec}^{2} (e) = 1 + { \tan}^{2} (e)[/tex]

It was given that,

[tex] { \tan}^{2} (e) = \frac{3}{8} [/tex]

We substitute the values into the identity to obtain,

[tex] { \sec}^{2} (e) = 1 + \frac{3}{8} [/tex]

[tex]{ \sec}^{2} (e) = \frac{11}{8} [/tex]

We take square root of both sides to get,

[tex]{ \sec}(e) = \pm \: \sqrt{\frac{11}{8} } [/tex]

Answer:

sec e = √(11/8) ⇒ answer B

Step-by-step explanation:

* Lets revise some identities in trigonometry

# sin²x + cos²x = 1

- Divide both sides by cos²x

∴ sin²x/cos²x + cos²x/cos²x = 1/cos²x

∵ sinx/cosx = tanx

∴ sin²x/cos²x = tan²x

∵ cos²x/cos²x = 1

∵ 1/cosx = secx

∴ 1/cos²x = sec²x

* Now lets write the new identity

# tan²x + 1 = sec²x

- Let x = e

∴ tan²e + 1 = sec²e

- Substitute the value of tan²e in the identity

∵ tan²e = 3/8

∴ 3/8 + 1 = sec²e

- Change the 1 to the fraction 8/8

∴ 3/8 + 8/8 = sec²e ⇒ add the fractions

∴ 11/8 = sec²e

- Take square root for the two sides to find sec e

∴ sec e = √(11/8)

∴ The answer is B

the answer is A. 110.5

The area of the rhombus is 110.5 square units if the lengths of the diagonal of a rhombus are 13 and 17 units option (A) 110.5 square units is correct.

A rhombus is a two-dimensional shape having four parallel opposite pairs of straight, equal sides. This shape resembles a diamond and is what you'd find on a deck of cards to represent the diamond suit. Rhombuses can be encountered in a variety of common situations.

We have a rhombus shown in the picture with dimensions:

As we know,

The area of the rhombus is given by:

A =pq/2

p and q are the lengths of the diagonal of a rhombus.

p = MK = 13 units

q = JL = 17 units

Plug the values in the formula:

A = (13×17)/2

A = 221/2

A = 110.5 square units

Thus, the area of the rhombus is 110.5 square units if the lengths of the diagonal of a rhombus are 13 and 17 units option (A) 110.5 square units is correct.

Learn more about the rhombus here:

brainly.com/question/22825947

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