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3 years ago

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Compare the average score of a student at the beginning of a month and at the end of that month, if at first, it increased by 40

%, then decreased by 25%

2 answers:

-BARSIC- [3]3 years ago

6 0

Answer: 5%

Step-by-step explanation:

Leno4ka [110]3 years ago

4 0

Answer:

122.5

Step-by-step explanation:

Let the student normal score be 100

In the beginning score increased by 40%

Hence score at the beginning =100+40 = 140%

Now before end score decreased by 25%

Decrease = 25% of 140 = 35

Hence score at the end of month = 140-35 = 105

Average score for the month

= Total of scores of end and beginning month /2

= $\frac{140+105}{2} \\=122.5$

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The graph shows the relationship between mass and volume for pure silver. Use the graph to determine the density of pure silver

Natalija [7]

Answer:

C. Slope and $10.49 g/cm^3$

Step-by-step explanation:

We're missing the graph itself, but I can make an educated guess:

- Assuming x = volume and y = mass:

$p = \frac{m}{v}$

density = mass/volume

x = volume = run

y = mass = rise

mass/volume = rise/run = slope

The slope of the line (assuming the graph is purely linear) represents mass/volume, which is the formula for density.

I had to Google the density of silver itself. Do the math from the graph for a more appropriate answer, as the graph may be less scientifically precise.

7 0

3 years ago

I need the answers to this question

erastovalidia [21]

Answer: <u>Madison can travel 2.2 miles less than Anthony on one gallon of gas.</u>

Step-by-step explanation:

The equation y = 39.1 $x$ represents the number of miles, $y$ , that madison can drive her car for every x gallons of gas.

Question: Madison can travel _ miles _ than Anthony on one gallon of gas.

Lets find how much Madison travels on 1 gallon of gas.

39.1(1) = 39.1

Therefore, Madison travels 39.1 miles on one gallon of gas.

Now lets solve for Anthony

Anthony travels 380.7 miles on 9 gallons of gas.

Therefore we can set up the equation:

9x = 380.7 to show that 9 gallons equates to 380.7 miles

to solve divide 9 from both sides

$\frac{9x}{9} =\frac{380.7}{9}$

x = 41.3

So, Anthony travels 41.3 miles on one gallon of gas

Now find whether Madison travels more or less than Anthony on one gallon.

39.1 - 41.3 = -2.2

This means that Anthony can travel 2.2 miles more than Madison.

For your answer, this means that:

<u>Madison can travel 2.2 miles less than Anthony on one gallon of gas</u>

<u></u>

<em>Hope this helped and have a nice day :)</em>

3 0

2 years ago

Really need help with this .

lions [1.4K]

Answer:

Step-by-step explanation:

The attached photo shows the diagram of quadrilateral QRST with more illustrations.

Line RT divides the quadrilateral into 2 congruent triangles QRT and SRT. The sum of the angles in each triangle is 180 degrees(98 + 50 + 32)

The area of the quadrilateral = 2 × area of triangle QRT = 2 × area of triangle SRT

Using sine rule,

q/SinQ = t/SinT = r/SinR

24/sin98 = QT/sin50

QT = r = sin50 × 24.24 = 18.57

Also

24/sin98 = QR/sin32

QR = t = sin32 × 24.24 = 12.84

Let us find area of triangle QRT

Area of a triangle

= 1/2 abSinC = 1/2 rtSinQ

Area of triangle QRT

= 1/2 × 18.57 × 12.84Sin98

= 118.06

Therefore, area of quadrilateral QRST = 2 × 118.06 = 236.12

6 0

3 years ago

Read 2 more answers

Show by substitution whether the number r is a solution of the corresponding quadratic equation.

Stels [109]

Answer:

It's choice B.

Step-by-step explanation:

When r= -11,

x^2 + 8x - 33 = (-11)^2 - 88 - 33

= 121 - 88 - 33

= 0

- so x = -11 gives a zero value and is a solution to the equation.

3 0

2 years ago

A simple random sample of items resulted in a sample mean of . The population standard deviation is . a. Compute the confidence

Varvara68 [4.7K]

Answer:

(a): The 95% confidence interval is (46.4, 53.6)

(b): The 95% confidence interval is (47.9, 52.1)

(c): Larger sample gives a smaller margin of error.

Step-by-step explanation:

Given

$n = 30$ -- sample size

$\bar x = 50$ -- sample mean

$\sigma = 10$ --- sample standard deviation

Solving (a): The confidence interval of the population mean

Calculate the standard error

$\sigma_x = \frac{\sigma}{\sqrt n}$

$\sigma_x = \frac{10}{\sqrt {30}}$

$\sigma_x = \frac{10}{5.478}$

$\sigma_x = 1.825$

The 95% confidence interval for the z value is:

$z = 1.960$

Calculate margin of error (E)

$E = z * \sigma_x$

$E = 1.960 * 1.825$

$E = 3.577$

The confidence bound is:

$Lower = \bar x - E$

$Lower = 50 - 3.577$

$Lower = 46.423$

$Lower = 46.4$ --- approximated

$Upper = \bar x + E$

$Upper = 50 + 3.577$

$Upper = 53.577$

$Upper = 53.6$ --- approximated

<em>So, the 95% confidence interval is (46.4, 53.6)</em>

Solving (b): The confidence interval of the population mean if mean = 90

First, calculate the standard error of the mean

$\sigma_x = \frac{\sigma}{\sqrt n}$

$\sigma_x = \frac{10}{\sqrt {90}}$

$\sigma_x = \frac{10}{9.49}$

$\sigma_x = 1.054$

The 95% confidence interval for the z value is:

$z = 1.960$

Calculate margin of error (E)

$E = z * \sigma_x$

$E = 1.960 * 1.054$

$E = 2.06584$

The confidence bound is:

$Lower = \bar x - E$

$Lower = 50 - 2.06584$

$Lower = 47.93416$

$Lower = 47.9$ --- approximated

$Upper = \bar x + E$

$Upper = 50 + 2.06584$

$Upper = 52.06584$

$Upper = 52.1$ --- approximated

<em>So, the 95% confidence interval is (47.9, 52.1)</em>

Solving (c): Effect of larger sample size on margin of error

In (a), we have:

$n = 30$ $E = 3.577$

In (b), we have:

$n = 90$ $E = 2.06584$

<em>Notice that the margin of error decreases when the sample size increases.</em>

4 0

3 years ago