All categories

3 years ago

X intercept for 2x +5y = -6

Mathematics

2 answers:

Vlad1618 [11]3 years ago

8 0

2x + 5y = -6
-5y -5y
___________
2x = -11y
__ ___
2 2
x = 5.5

Anettt [7]3 years ago

5 0

Answer:

(-3, 0)

Step-by-step explanation:

The x-intercept is the value of x when y = 0.

2x + 5y = -6

Set y = 0 2x + 5(0) = -6

2x = -6

Divide each side by 2 x = -3

The x-intercept is at (-3, 0).

You might be interested in

23/50 of Zanders friends prefer to stay home on Friday night what percent of Xander's friends prefer to stay home on Friday nigh

koban [17]

Answer:

46% of Xander's friends prefer to stay home on Friday nights.

Step-by-step explanation:

Formula

$Percentage = \frac{Part\ value\times 100}{Total\ value}$

As given

$\frac{23}{50}\ of\ Zanders\ friends\ prefer\ to\ stay\ home\ on\ Friday\ night.$

i.e

Zanders friends prefer to stay home on Friday night = 23

Total numbers of Zanders friends are 50.

Part value = 23

Total value = 50

Put in the formula

$Percentage = \frac{23\times 100}{50}$

$Percentage = \frac{2300}{50}$

Percentage = 46%

Therefore 46% of Xander's friends prefer to stay home on Friday nights.

3 0

3 years ago

Conjunction and disconjuction

maks197457 [2]

Step-by-step explanation:

When two statements are connected with an 'and,' you have a conjunction.

For conjunctions, both statements must be true for the compound statement to be true.

When the connector between two statements is "or," you have a disjunction. In this case, only one statement in the compound statement needs to be true for the entire compound statement to be true.

7 0

2 years ago

I need help with number 12

mestny [16]

(r - s)³ + r²
= (-3 - (-4))³ + (-3)²
= (-3 + 4)³ + 9
= 1³ + 9
= 1 + 9
= 10

6 0

2 years ago

3) What is the mean value of the data?

soldi70 [24.7K]

The mean - add all the numbers (data) divided by the amount. 4+3+1+1+2+4+1 =16 16/7 = 2.28
The median - the middle number.
The mode - the most common number.

8 0

2 years ago

A cellular phone company monitors monthly phone usage. The following data represent the monthly phone use in minutes of one part

Sergeu [11.5K]

Answer:

The standard deviation increased but there was no change in the interquantile range

Step-by-step explanation:

We are given the following data in the question:

320, 411, 348, 537, 420, 449, 462, 403, 454, 517, 515, 358, 438, 541, 387, 368, 502, 437, 431, 428.

n = 20

a) Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{8726}{20} = 436.3$

Sum of squares of differences =

13525.69 + 640.09 + 7796.89 + 10140.49 + 265.69 + 161.29 + 660.49 + 1108.89 + 313.29 + 6512.49 + 6193.69 + 6130.89 + 2.89 + 10962.09 + 2430.49 + 4664.89 + 4316.49 + 0.49 + 28.09 + 68.89 = 75924.2

$S.D = \sqrt{\frac{75924.2}{19}} = 63.21$

Sorted Data = 320, 348, 358, 368, 387, 403, 411, 420, 428, 431, 437, 438, 449, 454, 462, 502, 515, 517, 537, 541

$IQR = Q_3 - Q_1\\Q_3 = \text{upper median},\\Q_1 = \text{ lower median}$

$Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}$

$Median = \frac{431 + 437}{2} = 434$

$Q_1 = \frac{387 + 403}{2} = 395\\\\Q_3 = \frac{462 + 502}{2} = 482$

IQR = $482 - 395 = 87$

b) After changing the observation

0, 411, 348, 537, 420, 449, 462, 403, 454, 517, 515, 358, 438, 541, 387, 368, 502, 437, 431, 428

$Mean =\displaystyle\frac{8406}{20} = 420.3$

Sum of squares of differences =

176652.09 + 86.49 + 5227.29 + 13618.89 + 0.09 + 823.69 + 1738.89 + 299.29 + 1135.69 + 9350.89 + 8968.09 + 3881.29 + 313.29 + 14568.49 + 1108.89 + 2735.29 + 6674.89 + 278.89 + 114.49 + 59.29 = 247636.2

$S.D = \sqrt{\frac{247636.2}{19}} = 114.16$

Sorted Data = 0, 348, 358, 368, 387, 403, 411, 420, 428, 431, 437, 438, 449, 454, 462, 502, 515, 517, 537, 541

$Median = \frac{431 + 437}{2} = 434$

$Q_1 = \frac{387 + 403}{2} = 395\\\\Q_3 = \frac{462 + 502}{2} = 482$

IQR = $482 - 395 = 87$

Thus. the standard deviation increased but there was no change in the interquantile range.

5 0

3 years ago