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3 years ago

7th grade math-please help!

Mathematics

2 answers:

BlackZzzverrR [31]3 years ago

4 0

a.false

b.true

c.true (98.75)

d.true

e.false

Hope Your Thanksgiving Goes Well, Here's A Turkey

-TheKoolKid1O1

wel3 years ago

4 0

DCMS? ............................. (it needed to be 20 characters long XD)

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I NEED HELP ASAP... PLEASE HELP

Evgen [1.6K]

The first one is y=6(x-2)^2+4 and the second one is y=-3/2(x+6)^2+4

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

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

A ticket to a movie theatre for a student is $7. The cost for an adult is $2 more than for a student. How much would it cost 13

MakcuM [25]

Answer: it would cost a total of $194

Step-by-step explanation:

A ticket to a movie theatre for a student is $7. The cost for an adult is $2 more than for a student.

If student ticket costs $x, adult ticket would cost x + 2

This means that the cost for an adult is

7 + 2 = $9

The cost of 13 adult tickets to the movie theatre would be

13 × 9 = $117.

The cost of 11 student tickets to the movie theatre would be

11 × 7 = $77

Total cost of 13 adult tickets and 11 student tickets would be

117 + 77 = $194

3 0

3 years ago

Hana buys a new dress for $40. how much does she pay if there is a 7% tax

Rzqust [24]

Answer:

$42.80

Step-by-step explanation:

We can set up an equation to solve this as such

$40*1.07=42.80$

6 0

3 years ago

2. In the figure, what is m ZC? (1 point)<br> 15°<br> 35°<br> 50°<br> 65°

Doss [256]

Answer:

35°

Step-by-step explanation:

By the property of intersecting secants outside of a circle.

$m\angle C = \frac{1}{2} \times (100° - 30°)\\\\m\angle C = \frac{1}{2} \times 70°\\\\\huge \red {\boxed {m\angle C = 35°}}$

3 0

3 years ago