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3 years ago

The square pool was turned into a rectangular one and it's area was enlarged by a factor of 5 by enlarging one side of the pool

by 3 meters and the other side by 14. What are the new dimensions of the pool?

Mathematics

1 answer:

ozzi3 years ago

7 0

$\boxed{ \tt{the \: dimensions \: are : }}\\ \boxed{ \tt{b = 9 \: meters} }\\ \boxed{ \tt{l = 20 \: meters}}$

$5 {s}^{2} = (3 + s) \times (14 + s) \\ 5 {s}^{2} = {s}^{2} + 17s + 42 \\ {4s}^{2} - 17s - 42 = 0 \\ s = \frac{ - ( - 17) + \sqrt{ {( - 17)}^{2} - 4(4)(42)} }{2(4)} \\ s = \frac{17 + 31}{8} \\ s = 6 \\ hence \to \:the \: dimensions \: are : \\ b = s + 3 = 6 + 3 = 9 \: meters \\ l = s + 14 = 6 + 14 = 20 \: meters$

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anastassius [24]

Answer:

7/33 = 21 repeating and I'm pretty sure the question wants you to write it as an improper fraction. so your answer would be 40/33

Step-by-step explanation:

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2 years ago

How many different digits are needed for the base 6 system?

lorasvet [3.4K]

Answer:

0, 1, 2, 3, 4 and 5

Step-by-step explanation:

In a nutshell, base six is a number system that employs the numbers 0 through 5 instead of 0 through 9 in each digit. As a result, a number that would be stated as 6 in base ten is written as 10. We count one, two, three, four, five, ten, eleven, twelve, thirteen, fourteen, fifteen, twenty, and so on.

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3 years ago

The College Board SAT college entrance exam consists of three parts: math, writing and critical reading (The World Almanac 2012)

Wittaler [7]

Answer:

Yes, there is a difference between the population mean for the math scores and the population mean for the writing scores.

Test Statistics = $\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$ .

Step-by-step explanation:

We are provided with the sample data showing the math and writing scores for a sample of twelve students who took the SAT ;

Let A = Math Scores ,B = Writing Scores and D = difference between both

So, $\mu_A$ = Population mean for the math scores

$\mu_B$ = Population mean for the writing scores

Let $\mu_D$ = Difference between the population mean for the math scores and the population mean for the writing scores.

Null Hypothesis, $H_0$ : $\mu_A = \mu_B$ or $\mu_D$ = 0

Alternate Hypothesis, $H_1$ : $\mu_A \neq \mu_B$ or $\mu_D \neq$ 0

Hence, Test Statistics used here will be;

$\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$ where, Dbar = Bbar - Abar

$s_D$ = $\sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}$

n = 12

Student Math scores (A) Writing scores (B) D = B - A

1 540 474 -66

2 432 380 -52

3 528 463 -65

4 574 612 38

5 448 420 -28

6 502 526 24

7 480 430 -50

8 499 459 -40

9 610 615 5

10 572 541 -31

11 390 335 -55

12 593 613 20

Now Dbar = Bbar - Abar = 489 - 514 = -25

Bbar = $\frac{\sum B_i}{n}$ = $\frac{474+380+463+612+420+526+430+459+615+541+335+613}{12}$ = 489

Abar = $\frac{\sum A_i}{n}$ = $\frac{540+432+528+574+448+502+480+499+610+572+390+593}{12}$ = 514

∑ $D_i^{2}$ = 22600 and $s_D$ = $\sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}$ = $\sqrt{\frac{22600 - 12*(-25)^{2} }{12-1} }$ = 37.05

So, Test statistics = $\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$

= $\frac{-25 - 0}{\frac{37.05}{\sqrt{12} } }$ follows $t_1_1$ = -2.34

Now at 5% level of significance our t table is giving critical values of -2.201 and 2.201 for two tail test. Since our test statistics doesn't fall between these two values as it is less than -2.201 so we have sufficient evidence to reject null hypothesis as our test statistics fall in the rejection region .

Therefore, we conclude that there is a difference between the population mean for the math scores and the population mean for the writing scores.

8 0

3 years ago

Please!!! Help!!! Brainliest of Right! it goes as 50 point in the end but i used 100 point if you help me i will help you!!

vekshin1

Answer:

2. 3.913 kg (3 dp)

3. light cream

4. 240 CoffeeStops

5. 7 CoffeeStops per square mile

6. 2,861 cups of coffee each day

Step-by-step explanation:

Given:

Skim milk density at 20 °C = 1.033 kg/l
Light cream density at 20 °C = 1.012 kg/l
1 liter = 0.264 gallons

Question 2

$\begin{aligned}\textsf{1 gallon} & = \sf \dfrac{1}{0.264}\:liters\\\\\implies \textsf{Mass (1 gallon of skim milk)} & = \sf Density \times Volume\\& = \sf 1.033\:kg/l \times \dfrac{1}{0.264}\:l\\& = \sf 3.913\:kg\:(3\:dp)\end{aligned}$

Therefore, the mass of 1 gallon of skim milk is 3.913 kg (3 dp)

---------------------------------------------------------------------------------------------

Question 3

Given:

Volume of liquid = 9 liters
Mass of liquid = 9.108 kg

$\begin{aligned}\implies \sf Density & = \sf \dfrac{Mass}{Volume}\\\\& = \sf \dfrac{9.108\:kg}{9\:l}\\\\& = \sf 1.012\:kg/l \end{alilgned}$

Therefore, the container holds light cream.

---------------------------------------------------------------------------------------------

Question 4

Given:

15 CoffeeStops per 100,000 people
Population of Manhattan ≈ 1,602,000 people

$\begin{aligned}\implies \textsf{Number of Coffeestops} & = \sf \dfrac{population}{density}\\\\& = \sf \dfrac{1,602,000}{100,000/15}\\\\& = \sf \dfrac{1,602,000}{100,000} \times 15\\\\& = \sf 240.3\end{aligned}$

Therefore, there are 240 CoffeeStops.

---------------------------------------------------------------------------------------------

Question 5

Given

Manhattan ≈ 34 square miles

$\begin{aligned}\implies \textsf{CoffeeStops density} & = \sf \dfrac{number\:of\:stores}{land\:area}\\\\& = \sf \dfrac{240}{34}\\\\& \approx \sf 7 \: \textsf{CoffeeStops per square mile}\end{aligned}$

Therefore, the density of CoffeeStops is 7 per square mile.

---------------------------------------------------------------------------------------------

Question 6

Given:

Each person buys 3 cups of coffee per week

$\begin{aligned}\implies \textsf{Cups served each week} & = \textsf{number of people} \times \textsf{number of cups per week}\\& = \sf 1,602,000 \times 3\\& = \sf 4,806,000\: \textsf{cups per week}\\\\\implies \textsf{Cups per day} & = \sf \dfrac{\textsf{cups per week}}{\textsf{days in a week}}\\\\& = \sf \dfrac{4,806,000}{7}\\\\& = \sf 686,571\:\textsf{(nearest whole number)}\end{aligned}$

$\begin{aligned}\implies \textsf{Cups served per day per shop} & = \dfrac{\textsf{cups per day}}{\textsf{number of shops}}\\\\& = \sf \dfrac{686,571}{240}\\\\& = \sf 2,861\: \textsf{(nearest whole number)} \end{aligned}$

Therefore, each Manhattan CoffeeStop serves approximately 2,861 cups of coffee each day.

7 0

2 years ago

7) Solve the following system of equations below algebraically using substitution.

Pani-rosa [81]

Answer:

x = 3.25

y =4.75

Step-by-step explanation:

In order to Solve the following system of equations below algebraically using substitution method we say that;

let;

8x - 4y = 7 ..................... equation 1

x + y = 8.......................... equation 2

from equation2

x + y = 8.......................... equation 2

x = 8 - y.............................. equation 3

substitute for x in equation 1

8x - 4y = 7 ..................... equation 1

8(8-y) - 4y = 7

64-8y-4y=7

64-12y=7

collect the like terms

64-7 = 12y

57= 12y

divide both sides by the coefficient of y which is 12

57/12 = 12y/12

4.75 = y

y =4.75

put y = 4.75 in equation 3

x = 8 - y.............................. equation 3

x = 8 -4.75

x = 3.25

to check if your answer is correct, put the value of x and y in either equation 1 or 2

from equation 2

x + y = 8.......................... equation 2

3.25 + 4.75 =8

8=8.................... proved

8 0

3 years ago