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Alexeev081 [22]
3 years ago
11

Find three points of the linear equation -x-2y=-10

Mathematics
1 answer:
n200080 [17]3 years ago
6 0
We can find 3 points by plugging them in

x = -12 y = 1
x = 10 y = 0
x = 0 y = 5
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Solve the given linear Diophantine equation. A) 12x+9y=15 B) 8x+10y=17
Nookie1986 [14]

Answer:

  A) (x, y) = (2-3n, 4n-1) . . . . for any integer n

  B) no solution

Step-by-step explanation:

A) All coefficients have a common factor of 3, so any solution of the reduced equation will be a solution of the given equation. The reduced equation is ...

  4x +3y = 5

A graph of the original shows (x, y) = (2, -1) is a solution. Then other solutions will be those values with a multiple of 4 added to y and the same multiple of 3 subtracted from x:

  (x, y) = (2 -3n, 4n -1)

__

B) The left side of the equation is an even number for any integer values of x and y. The right side is an odd number. There can be no solution.

5 0
3 years ago
Will positive numbers always have a higher absolute value than negative numbers?
Reika [66]
-I5I=-5 vs. I-5I= 5

The absolute value will stay the same for both negative and positive. The result will change depending on where you put your absolute value signs.
6 0
3 years ago
• A researcher claims that less than 40% of U.S. cell phone owners use their phone for most of their online browsing. In a rando
antiseptic1488 [7]

Answer:

We failed to reject H₀

Z > -1.645

-1.84 > -1.645

We failed to reject H₀

p > α

0.03 > 0.01

We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

Step-by-step explanation:

Set up hypotheses:

Null hypotheses = H₀: p = 0.40

Alternate hypotheses = H₁: p < 0.40

Determine the level of significance and Z-score:

Given level of significance = 1% = 0.01

Since it is a lower tailed test,

Z-score = -2.33 (lower tailed)

Determine type of test:

Since the alternate hypothesis states that less than 40% of U.S. cell phone owners use their phone for most of their online browsing, therefore we will use a lower tailed test.

Select the test statistic:  

Since the sample size is quite large (n > 30) therefore, we will use Z-distribution.

Set up decision rule:

Since it is a lower tailed test, using a Z statistic at a significance level of 1%

We Reject H₀ if Z < -1.645

We Reject H₀ if p ≤ α

Compute the test statistic:

$ Z =  \frac{\hat{p} - p}{ \sqrt{\frac{p(1-p)}{n} }}  $

$ Z =  \frac{0.31 - 0.40}{ \sqrt{\frac{0.40(1-0.40)}{100} }}  $

$ Z =  \frac{- 0.09}{ 0.048989 }  $

Z = - 1.84

From the z-table, the p-value corresponding to the test statistic -1.84 is

p = 0.03288

Conclusion:

We failed to reject H₀

Z > -1.645

-1.84 > -1.645

We failed to reject H₀

p >  α

0.03 > 0.01

We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

8 0
3 years ago
Two square are congruent ,if they have same ______​
OverLord2011 [107]

Two squares are congruent if they have the same side length.

3 0
3 years ago
Read 2 more answers
Karen spends $450 on monthly bills. Of this total amount, 12% is for phone service, 1/10 is for internet service, and 2/9 is for
finlep [7]
12%= 0.12 = 12/100 = 108/900
1/10 = 90/900
2/9 = 200/900

108/900 + 90/900 + 200/900 = 398/900 used

900- 398 = 502

502/900 = 251/ 450

251/450 × 450 = 251


Karen has $251 for food
4 0
2 years ago
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