All categories

3 years ago

I WILL MARK BRAINLIEST PLEASE HURRY

Mathematics

1 answer:

AfilCa [17]3 years ago

4 0

Consider, pls, this explanation + solution:
according to the given equation the centre of this circle is in the poit (6;7) and its radius =4 units.
Using this data, it is possible to define right choice: D.

answer: graph D.

You might be interested in

What is the answer to 8w+56?

adelina 88 [10]

Answer:

What is w equal to?

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Please help ASAP ASAP please

Serhud [2]

Answer:

PT = 1.5 , VT = 1.5√3

Step-by-step explanation:

Given

PV = 3

With reference angle 60°

hypotenuse (h) = 3

perpendicular (p) = ?

sin 60° = p/h

√3 / 2 = p / 3

p = √3 * 3 / 2

p = 1.5√3

With reference angle 30°

hypotenuse (h) = 3

perpendicular (p) = ?

sin 30° = p/h

1/2 = p/ 3

p = 1.5

Hope it will help :)❤

3 0

3 years ago

Question 9 (1 point)

Romashka [77]

Answer:

Step-by-step explanation:

League A League B

151.12 163.25

148 157

26.83 24.93

29 136

136 145

167 178

207 256

League A in ascending order :

26.83 , 29 , 136, 148 , 151.12 , 167,207

$Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\\\Mean = \frac{26.83+29 +136+ 148+ 151.12+ 167+207}{7}\\\\Mean =123.564$

Median = Mid value of data

n = 7

So, mid value = 4th term

Median=148

Standard deviation= $\sqrt{\frac{\sum(x_i-\bar{x})^2}{n}}$ = $=\sqrt{\frac{(26.83-123.564)^2+(29-123.564)^2+.......+(207-123.564)^2}{7}}=63.98$

To Find Q1

Q1 is the mid value of lower quartile

Lower quartile : 26.83 , 29 , 136, 148

n = 4

Q1=82.5

To Find Q3

Q3 is the mid value of upper quartile

Upper quartile : 148 , 151.12 , 167,207

n = 4

Q3=159.06

IQR = Q3-Q1=159.06-82.5=76.56

To find outlier

(Q1-1.5IQR ,Q3+1.5IQR)

$(82.5-1.5\times 76.56,159.06+1.5\times 76.56)$

(-32.34,273.9)

So, There is no outlier

Maximum = 207

League B in ascending order :

24.93,136,145,157,163.25,178,256

$Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\\\Mean = \frac{24.93+136+145+157+163.25+178+256}{7}\\\\Mean =151.45$

Median = Mid value of data

n = 7

So, mid value = 4th term

Median=157

Standard deviation= $\sqrt{\frac{\sum(x_i-\bar{x})^2}{n}}$ = $=\sqrt{\frac{(24.93-151.45)^2+(136-151.45)^2+.......+(256-151.45)^2}{7}}=68.42$

To Find Q1

Q1 is the mid value of lower quartile

Lower quartile : 24.93,136,145,157

n = 4

$Median = \frac{\frac{n}{2} \text{th term}+(\frac{n}{2}+1) \text{th term}}{2}\\Median = \frac{\frac{4}{2} \text{th term}+(\frac{4}{2}+1) \text{th term}}{2}\\Median = \frac{2 \text{th term}+3 \text{th term}}{2}\\Median = \frac{136+145}{2}=140.5$

Q1=140.5

To Find Q3

Q3 is the mid value of upper quartile

Upper quartile : 157,163.25,178,256

n = 4

Q3=170.625

IQR = Q3-Q1=170.625-140.5=30.125

To find outlier

(Q1-1.5IQR ,Q3+1.5IQR)

$(140.5-1.5\times 30.125,170.625+1.5\times 30.125)$

(95.3125,215.8125)

24.93 and 256 are outliers

Maximum = 256

5 0

3 years ago

5x+2y = 19<br> 1x + 4y=13

Mama L [17]

Answer:

{x,y} = {25/9,23/9}

Step-by-step explanation:

// Solve equation [2] for the variable x

[2] x = -4y + 13

// Plug this in for variable x in equation [1]

[1] 5•(-4y+13) + 2y = 19

[1] - 18y = -46

// Solve equation [1] for the variable y

[1] 18y = 46

[1] y = 23/9

// By now we know this much :

x = -4y+13

y = 23/9

// Use the y value to solve for x

7 0

3 years ago

Read 2 more answers

Steven wishes to save for his retirement by depositing $2,000 at the beginning of each year for thirty years. Exactly one year a

Ad libitum [116K]

Step-by-step explanation:

i = interest 3% for 30 years

This is a simple dynamical system for whom the the solutions are given as

$S=R[\frac{(i+1)^n-1}{i}](i+1)$

putting values we get

S=2000[\frac{(1.03)^{30}-1}{0.03}](1.03)

= $98005.35

withdrawal of money takes place from one year after last payment

To determine the result we use the present value formula of an annuity date

$P = R\frac{1-(1+i)^{-n}}{i}{i+1}$

we need to calculate R so putting the values and solving for R we get

R= $6542.2356

8 0

3 years ago