Please help me! What is the simplest form for 21 /63 ?

Calculate the slope of the line on the graph using the formula: m=y2-y1/x2-x1

Fudgin [204]

The slope is 4 m/s
Hope this helps

6 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

2)

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

At which root does the graph of f(x) = (x+4)6(x + 7)5 cross the x axis?

Licemer1 [7]

Answer:

x = -7

Step-by-step explanation:

Our function is $f(x) =(x+4)^6(x+7)^5$ . Notice that our possible roots are when x + 4 = 0 and when x + 7 = 0. So, our roots are -4 and -7.

However, the power above x + 4 is even, meaning the graph will simply <em>touch</em> the x-axis at x = -4, but not pass through. The power above x + 7, though, is odd, which means the graph will <em>cross</em> the x-axis.

Thus, the answer is x = -7.

5 0

3 years ago

Read 2 more answers

Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). To estimate

Leviafan [203]

Answer:

Probability of having student's score between 505 and 515 is 0.36

Given that z-scores are rounded to two decimals using Standard Normal Distribution Table

Step-by-step explanation:

As we know from normal distribution: z(x) = (x - Mu)/SD

where x = targeted value; Mu = Mean of Normal Distribution; SD = Standard Deviation of Normal Distribution

Therefore using given data: Mu (Mean) = 510, SD = 10.4 we have z(x) by using z(x) = (x - Mu)/SD as under:

In our case, we have x = 505 & 515

Approach 1 using Standard Normal Distribution Table:

z for x=505: z(505) = (505-510)/10.4 gives us z(505) = -0.48

z for x=515: z(515) = (515-510)/10.4 gives us z(515) = 0.48

Afterwards using Normal Distribution Tables and rounding the values to two decimals we find the probabilities as under:

P(505) using z(505) = 0.32

Similarly we have:

P(515) using z(515) = 0.68

Now we may find the probability of student's score between 505 and 515 using:

P(505 < x < 515) = P(515)-P(505) = 0.68 - 0.32 = 0.36

PS: The standard normal distribution table is being attached for reference.

Approach 2 using Excel or Google Sheets:

P(x) = norm.dist(x,Mean,SD,Commutative)

P(505) = norm.dist(505,510,10.4,1)

P(515) = norm.dist(515,510,10.4,1)

Probability of student's score between 505 and 515= P(515) - P(505) = 0.36

Download pdf

6 0

3 years ago

For the given bond, whose interest rate is provided, find the semiannual interest payment and the

Vika [28.1K]

Answer:

The semiannual interest payment is $485.5

The total interest earned over the life of the bond is $10,681

Step-by-step explanation:

The formula of the simple interest is I = Prt, where

P is the money invested
r is the interest rate in decimal
t is the time in years

∵ The bond is $20,000

∴ P = 20,000

∵ First Income Fund 11-year bond at 4.855%

∴ t = 11

∴ r = 4.855%

- Change it to decimal by divide it by 100

∴ r = 4.855 ÷ 100 = 0.04855

∵ The semiannual interest payment = P × r × half a year

∴ The semiannual interest payment = 20,000 × 0.04855 × $\frac{1}{2}$ (1)

∴ The semiannual interest payment = 485.5

The semiannual interest payment is $485.5

∵ I = Prt

- Substitute the values of P, r and t in the formula

∴ I = 20,000(0.04855)(11)

∴ I = 10,681

The total interest earned over the life of the bond is $10,681

6 0

3 years ago