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denis-greek [22]

4 years ago

7

Below, five systems of linear equations have been put in reduced row echelon form. Identify how many solutions each one has, and

enter that number in the blank. Enter the word "infinite" (without the quote marks) if there are infinitely many solutions. (a) 1 equation editorEquation Editor ⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢100000000000−400000010000050000∣∣∣∣∣∣∣∣∣−5−40000⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥ (b) 6 equation editorEquation Editor ⎡⎣⎢⎢⎢⎢⎢⎢1000001000001000001000001∣∣∣∣∣∣∣−4420−2⎤⎦⎥⎥⎥⎥⎥⎥ (c) 3 equation editorEquation Editor ⎡⎣⎢100−2000000100−10440∣∣∣∣401⎤⎦⎥ (d) 5 equation editorEquation Editor ⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢100000010000001000000100000010∣∣∣∣∣∣∣∣∣2−5−5020⎤⎦⎥⎥⎥⎥⎥⎥⎥⎥ (e) 3 equation editorEquation Editor ⎡⎣⎢⎢⎢⎢⎢⎢10000−2000001000−52000−3−4000∣∣∣∣∣∣∣3−4100⎤⎦⎥⎥⎥⎥⎥⎥

1 answer:

vesna_86 [32]4 years ago

8 0

Broken Calculator checkkkkkkk

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if you have 678 t-shirts and you want to put the same amount of t-shirts in 4 boxes how would you do that?

KIM [24]

In order to solve this problem, you will need to use a type of operation. You can do long division which will be 678 divided 4 and it will give you 169 remainder 5.<span />

4 0

4 years ago

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What is the slope of the line that passes through the points (2, 3)

Komok [63]

Slope is found when you put y2-y1/x2-x1. So your equation would be (-2)-2/6-3.
That answer would be -4/3. Nothing can be simplified further, so that is your answer

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3 years ago

Rearrange this to make a the subject

lyudmila [28]

Answer:

w = 3(2a + b) - 4

w = 6a + 3b - 4

a = (w - 3b + 4) / 6

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4 years ago

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In a random sample of 380 cars driven at low altitudes, 42 of them exceeded a standard of 10 grams of particulate pollution per

alexdok [17]

Answer:

The test statistic for testing if the proportion of high-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles exceeding the standard is 3.234.

Step-by-step explanation:

We are given that in a random sample of 380 cars driven at low altitudes, 42 of them exceeded a standard of 10 grams of particulate pollution per gallon of fuel consumed.

In an independent random sample of 90 cars driven at high altitudes, 24 of them exceeded the standard.

Let $p_1$ = <u><em>population proportion of cars driven at high altitudes who exceeded a standard of 10 grams</em></u>.

$p_2$ = <u><em>population proportion of cars driven at low altitudes who exceeded a standard of 10 grams</em></u>.

So, Null Hypothesis, $H_0$ : $p_1\leq p_2$ {means that the proportion of high-altitude vehicles exceeding the standard is smaller than or equal to the proportion of low-altitude vehicles exceeding the standard}

Alternate Hypothesis, $H_A$ : $p_1>p_2$ {means that the proportion of high-altitude vehicles exceeding the standard is greater than the proportion of low-altitude vehicles exceeding the standard}

The test statistics that will be used here is <u>Two-sample z-test statistics</u> for proportions;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of cars driven at high altitudes who exceeded a standard of 10 grams = $\frac{24}{90}$ = 0.27

$\hat p_2$ = sample proportion of cars driven at low altitudes who exceeded a standard of 10 grams = $\frac{42}{380}$ = 0.11

$n_1$ = sample of cars driven at high altitudes = 90

$n_2$ = sample of cars driven at low altitudes = 380

So, the test statistics = $\frac{(0.27-0.11)-(0)}{\sqrt{\frac{0.27(1-0.27)}{90}+\frac{0.11(1-0.11)}{380} } }$

= 3.234

The value of z-test statistics is 3.234.

7 0

3 years ago

A 5×6 rectangular grid with unit cells is given. Let A(0, 0) and B(5, 6) be the bottom-left corner and top-right corner, respect

harina [27]

Answer:

462 paths from A to B

Step-by-step explanation:

The question is incomplete. However, a possible question is to determine the number of possible paths on the grid map.

The first step is to represent the grid map itself. (See attachment 1)

From the question, we understand that:

Only right movement is allowed in the horizontal direction

Only up movement is allowed in the vertical direction

There are a several number of ways to navigate through. However, one possible way is in attachment 2

In attachment 2,

R represents the right movement
U represents the up movement

And we have:

$R = 5$ and $U = 6$

The number of possible paths (N) is then calculated as:

$N = \frac{(N + U)!}{N!U!}$

Substitute values for N and U

$N = \frac{(5 + 6)!}{5!6!}$

$N = \frac{11!}{5!6!}$

$N = \frac{39916800}{120 * 720}$

$N = \frac{39916800}{86400}$

$N = 462$

<em>Hence, there are 462 possible paths from A to B</em>

4 0

3 years ago