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3 years ago

9

PLZ HELP QUICK I NEED TO TURN THIS IN ;Twice a number is fourteen. Write and solve an equation to nd the number of miles that mu

st be run each
week.

1 answer:

LuckyWell [14K]3 years ago

5 0

2n=14 n being the “number” divide 2 on both sides and you’re left w n=14/2 or7

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What should be subtracted from 5/8 to make it -1

user100 [1]

Answer:

13/8

Step-by-step explanation:

let the number be x

5/8 - x = -1

-x = -1 - 5/8

-x = - 13/8

x = 13/8

5/8- 13/8 = -1

5 0

2 years ago

Read 2 more answers

A rectangular ink pad has a permieter of 26 cenimeters. Ts area is 40 square cenimeters. What are the deminitions of the ink pad

Assoli18 [71]

Answer: 5cm and 8cm

Step-by-step explanation:

Perimeter of a rectangle = 2l + 2w

Area of a rectangle = l × w

where l = length

w = width

Therefore,

2l + 2w = 26 ...... i

l × w = 40 ...... ii

We can reduce equation i to l + w = 13.

l + w = 13 ....... i

l × w = 40 ....... ii

From i, l = 13 - w ...... iii

Put the value of l into equation ii

l × w = 40

(13 - w) × w = 40

13w - w² = 40

13w - w² - 40 = 0

w² - 13w + 40 = 0

w² -8w -5w +40 = 0

w(w - 8) -5(w - 8) = 0

(w - 5) = 0 and (w - 8) = 0

w = 0 +5 and w = 0 + 8

w = 5

l = 8

The dimensions are 5cm and 8cm

8 0

3 years ago

What number has 7 ten thousands, 1 thousand, 1 hundred, and no ones?

FrozenT [24]

Answer:

$71,100$

Step-by-step explanation:

If you are trying to find a number that is written in word form, we can just use place values to find what goes where.

A number is broken down into this:

Ten thousands, thousands, hundreds, tens, ones.

If they have 7 ten thousands, the first digit will be a 7.

If they have 1 thousand, the second digit will be a 1.

If they have 1 hundred, the third digit will be a 1.

Since nothing is stated about tens, we assume it's value is 0.

And since there are no ones, it's value is 0.

So:

71,100.

Hope this helped!

8 0

4 years ago

The ages of students enrolled in two math classes at the local community college, Class A and Class B, are listed in order below

nlexa [21]

Answer:

The true statement about Class B is that Class B has a smaller median and the same inter quartile range.

Step-by-step explanation:

We are given the ages of students enrolled in two math classes at the local community college, Class A and Class B, below;

Class A: 20, 20, 20, 21, 22, 23, 23, 25, 27, 29, 30, 31, 34, 35, 36, 39, 40

Class B: 16, 17, 18, 18, 20, 22, 22, 24, 26, 26, 28, 29, 30, 34, 37, 40, 42

1) <u>Firstly, we will calculate Median for Class A;</u>

For calculating median, first we have to observe that number of observations (n) in our data is even or odd, that is;

If n is odd, then the formula for calculating median is given by;

Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

If n is even, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th} \text{ obs.}+(\frac{n}{2}+1)^{th} \text{ obs.}}{2}$

Here, number of observation is odd, i.e. n = 17.

So, Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

= $(\frac{17+1}{2})^{th} \text{ obs.}$

= $(\frac{18}{2})^{th} \text{ obs.}$

= $9^{th} \text { obs.}$ = 27

Hence, the median of class A is 27.

2) <u>Now, we will calculate Median for Class B;</u>

For calculating median, first we have to observe that number of observations (n) in our data is even or odd, that is;

If n is odd, then the formula for calculating median is given by;

Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

If n is even, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th} \text{ obs.}+(\frac{n}{2}+1)^{th} \text{ obs.}}{2}$

Here, number of observation is odd, i.e. n = 17.

So, Median = $(\frac{n+1}{2})^{th} \text{ obs.}$

= $(\frac{17+1}{2})^{th} \text{ obs.}$

= $(\frac{18}{2})^{th} \text{ obs.}$

= $9^{th} \text { obs.}$ = 26

Hence, the median of class B is 26.

3) Now, we will calculate the Inter quartile range for Class A;

Inter quartile range = Upper quartile - Lower quartile

= $Q_3-Q_1$

SO, $Q_1 = (\frac{n+1}{4})^{th} \text{ obs.}$

= $(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{18}{4})^{th} \text{ obs.}$

= $4.5^{th} \text{ obs.}$

= $4^{th} \text{ obs.} + 0.5[5^{th} \text{ obs.} - 4^{th} \text{ obs.}]$

= $21+ 0.5[22- 21]$

= 21.5

Similarly, $Q_3 = 3(\frac{n+1}{4})^{th} \text{ obs.}$

= $3(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{54}{4})^{th} \text{ obs.}$

= $13.5^{th} \text{ obs.}$

= $13^{th} \text{ obs.} + 0.5[14^{th} \text{ obs.} - 13^{th} \text{ obs.}]$

= $34+ 0.5[35- 34]$

= 34.5

Therefore, Inter quartile range for Class A = 34.5 - 21.5 = 13.

4) Now, we will calculate the Inter quartile range for Class B;

Inter quartile range = Upper quartile - Lower quartile

= $Q_3-Q_1$

SO, $Q_1 = (\frac{n+1}{4})^{th} \text{ obs.}$

= $(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{18}{4})^{th} \text{ obs.}$

= $4.5^{th} \text{ obs.}$

= $4^{th} \text{ obs.} + 0.5[5^{th} \text{ obs.} - 4^{th} \text{ obs.}]$

= $18+ 0.5[20- 18]$

= 19

Similarly, $Q_3 = 3(\frac{n+1}{4})^{th} \text{ obs.}$

= $3(\frac{17+1}{4})^{th} \text{ obs.}$

= $(\frac{54}{4})^{th} \text{ obs.}$

= $13.5^{th} \text{ obs.}$

= $13^{th} \text{ obs.} + 0.5[14^{th} \text{ obs.} - 13^{th} \text{ obs.}]$

= $30+ 0.5[34- 30]$

= 32

Therefore, Inter quartile range for Class B = 32 - 19 = 13.

Hence, the true statement about Class B is that Class B has a smaller median and the same inter quartile range.

4 0

4 years ago

What else would need to be congruent to show that triangle ABC and triangle XYZ is congruent by AAS

julia-pushkina [17]

Answer:

C. ∠B ≅ ∠Y

Step-by-step explanation:

This is because AAS means 2 angles and 1 side are congruent.

Which means there are two possible answers: B or C. But answer B says angle C and Z are congruent which would make both triangles congruent by ASA and not AAS

7 0

3 years ago