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3 years ago

For what values of a can you solve the linear system ax+3y=2 and 4x+5y=6 by elimination without multiplying first?

Mathematics

1 answer:

coldgirl [10]3 years ago

5 0

Answer:

a=-4

Step-by-step explanation:

$ax+3y=2$ and $4x+5y=6$

In elimination method we try to eliminate one variable

To eliminate one variable we need to make the coefficients same with different sign

to eliminate the variable x we need to make the number before x same and with different sign

In the second equation the coefficient of x is 4

so the value of 'a' should be -4 to eliminate

a=-4

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What is the average rate of change of f(x), represented by the graph, over the interval [-2, 2]?

Molodets [167]

To get the average rate of change (ARC) of f(x) over [x1, x2], we use the formula:
ARC = ( f(x2) - f(x2) ) / (x2 - x1)

From the graph
f(2) = 4
f(-2) = 4

Plugging in the values into the formula:
ARC = (4 - 4) / (2 - (-2) )
ARC = 0

The points connecting (-2,4) amd (2,4) is a horizontal line that is the rate of change is 0.

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Tamiku [17]

Answer:

Option C, F(x) = (x+2)/(x+1)

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3 years ago

If 20 is decreased to 18.6, the decrease is what percent of the original number?

Lady bird [3.3K]

Answer:

1.4

Step-by-step explanation:

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3 years ago

(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each

Leviafan [203]

Following are the solution parts for the given question:

For question A:

$\to (n) = 16$

$\to (\bar{X}) = 410$

$\to (\sigma) = 40$

In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

$\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}$

$\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$ When $90\%$ of the confidence interval:

$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

$\to (n) = 500$

$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

$\to C.I= 0.90$

$\to (\alpha) = 0.10$

$\to \frac{\alpha}{2} = 0.05$

Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:

brainly.in/question/16329412

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3 years ago

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Viktor [21]

Answer:

i think the answer is A. 14 surfaces with glue on them

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