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prohojiy [21]
3 years ago
10

give an example if a linear equation in one variable with infinitely many solutioms. Explain how you know it has many solutions.

Mathematics
2 answers:
scoray [572]3 years ago
6 0
Ask expert and try to use your brain and why do you want to know many solutions just study one that's enough
earnstyle [38]3 years ago
5 0
There can be zero solutions, 1 solution or infinite solutions--each case is explained in detail below. <span>Note: Although systems of linear equations can have 3 or more equations,we are going to refer to the most common case--a stem with exactly 2 lines.</span> Case I: 1 Solution

This is the most common situation and it involves lines that intersect exactly 1 time.

Case 2: No Solutions

This only happens when the lines are parallel. As you can see, parallel lines are not going to ever meet.

Example of a stem that has no solution:

<span><span>Line 1: y = 5x +13</span><span>Line 2: y = 5x + 12</span></span>
Case 3: Infinite Solutions

This is the rarest case and only occurs when you have the same line
Consider, for instance, the two lines below (y = 2x+1 and 2y = 4x +2). These two equations are really the same line .

Example of a system that has infinite solutions:

<span><span>Line 1: y = 2x + 1</span><span>Line 2: 2y = 4x + 2 </span></span>





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jonas has 8 sponsors for the school walk-a-thon. Maura has 3 times as many sponsors as Jonas. Treton has 1/4 as many sponsors as
kirill [66]
Jonas has 8, Maura has 24, and Treton has 6 I think. Hope it helps :)
3 0
3 years ago
Read 2 more answers
The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days.
Zarrin [17]
Let the lengths of pregnancies be X
X follows normal distribution with mean 268 and standard deviation 15 days
z=(X-269)/15

a. P(X>308)
z=(308-269)/15=2.6
thus:
P(X>308)=P(z>2.6)
=1-0.995
=0.005

b] Given that if the length of pregnancy is in lowest is 44%, then the baby is premature. We need to find the length that separates the premature babies from those who are not premature.
P(X<x)=0.44
P(Z<z)=0.44
z=-0.15
thus the value of x will be found as follows:
-0.05=(x-269)/15
-0.05(15)=x-269
-0.75=x-269
x=-0.75+269
x=268.78
The length that separates premature babies from those who are not premature is 268.78 days
8 0
3 years ago
Josie rode her bike 18 miles at 12 and then rode another 24 miles at 15.
sleet_krkn [62]
18/12= 1.5 
24/15= 1.6
1.6+1.5=3.1
I think D is your answer 
8 0
3 years ago
The following data represent weights (pounds) of a random sample of professional football players on the following teams. X1 = w
alexandr1967 [171]

I don't have an idea of what the answer is to be honest.

8 0
3 years ago
A researcher collected data on the hours of TV watched per day from a sample of five people of different ages. Here are the resu
frozen [14]

Answer:

1. The least squares regression is y = -0.1015·x + 6.51

2. The independent variable is b) age

Please see attached table

Step-by-step explanation:

The least squares regression formula is given as follows;

\dfrac{\sum_{i = 1}^{n}(x_{i} - \bar{x})\times \left (y_{i} - \bar{y}  \right ) }{\sum_{i = 1}^{n}(x_{i} - \bar{x})^{2}}

We have;

\bar x = 24

\bar y = 4

\Sigma (x_i - \bar x) (y_i - \bar y) = -79

\Sigma (x_i - \bar x)^2 = 778

\therefore \hat \beta =\dfrac{\sum_{i = 1}^{n}(x_{i} - \bar{x})\times \left (y_{i} - \bar{y}  \right ) }{\sum_{i = 1}^{n}(x_{i} - \bar{x})^{2}} = \frac{-79}{778} = -0.1015

The least squares regression is y = -0.1015·x + α

∴ α = y  -0.1015·x = 6 - (-0.1015 × 5) = 6.51

The least squares regression is thus;

y = -0.1015·x + 6.51

2. The independent variable is the age b)

3. Steps to create an ANOVA table with α = 0.05

The overall mean = (43  + 30  + 22  + 20  + 5  + 1  + 6  + 4  + 3  + 6 )/10 = 14

There are 2 different treatment = df_{treat} = 2 - 1 = 1

There are 10 different treatment measurement = df_{tot} = 10 - 1 = 9

df_{res} = 9 - 1 = 8

df_{treat} + df_{res} = df_{tot}

The estimated effects are;

\hat A_1 = 24 - 14 = 10

\hat A_2 = 4 - 14 = -10

SS_{treat} = 10^2 \times 5 + (-10)^2 \times 5 =1000

\sum_{i}\SS_{row}_i = \sum_{i}\sum_{j} (y_{ij} - \bar y)= [(1 - 4)^2 + (6 - 4)^2 + (4 - 4)^2 + (3 - 4)^2 + (6 - 4)^2] = 18

\sum_{i} S S_{row}_i = \sum_{i}\sum_{j} (y_{ij} - \bar y) ^2= [(43 - 24)^2 + (30 - 24)^2 + (22 - 24)^2 + (20 - 24)^2 + (5 - 24)^2] = 778

S S_{res} = \sum_{i} S S_{row}_i = 778 + 18 = 796

SS_{tot} = (43  - 14)² + (30  - 14)² + (22  - 14)² + (20  - 14)² + (5  - 14)² + (1 - 14)1² + (6  - 4 )² + (3  - 14)² + (6  - 14)² = 1796

MS_{treat} = \dfrac{SS_{treat} }{df_{treat} } = \dfrac{1000}{1} = 1000

MS_{res} = \dfrac{SS_{res} }{df_{res} } = \dfrac{796}{8} = 99.5

F- value is given by the relation;

F = \dfrac{MS_{treat} }{MS_{res} } = \frac{1000}{99.5} = 10.05

We then look for the critical values at degrees of freedom 1 and 8 at α = 0.05 on the F-distribution tables 5.3177

Hence; F = 10.05 > F_{1,8}^{Krit}(5\%) = 5.3177, we reject the null hypothesis.

7 0
3 years ago
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