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3 years ago

Solve the system by elimination:2x+3y=44x+6y=9

1 answer:

6 0

$\left \{ {2x+3y=4\ \ |*(-2)} \atop {4x+6y=9}} \right. \\\\ \left \{ {-4x-6y=-8} \atop {4x+6y=9}} \right. \\+------\\Elimination\ method\\\\
0 \neq 1\ \ \\This\ is\ not\ equal\ so\ there\ are\ no\ solutions.$

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Afina-wow [57]

Answer:

V≈278.85

Step-by-step explanation:

AB=s(s﹣a)(s﹣b)(s﹣c)

V=ABh

s=a+b+c

Solving forV

V=1

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3 years ago

Lin’s family has completed 70% of a trip. They have traveled 35 miles how far is the trip

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ANSWER:The trip is 50 miles.

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Read 2 more answers

7,421,273 in word form

lukranit [14]

Answer:

Seven million, four hundred and twenty-one thousand, two hundred and seventy-three

Step-by-step explanation:

7,421,273 in word form is seven million, four hundred and twenty-one thousand, two hundred and seventy-three 

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4 0

3 years ago

Pls helpi need it with this quiz

Bad White [126]

Answer:

Step-by-step explanation:

one dollar = $1

2 1/2 = 2.5

$1 / 2.5 = $0.40 or 40 cents

$3.78 / 6 = $0.63 or 63 cents

$1.56 / 20 = $0.078

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3 years ago

At Ajax Spring Water, a half-liter bottle of soft drink is supposed to contain a mean of 519 ml. The filling process follows a n

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Answer:

1) The normal distribution should be used for the sample mean because the population distribution is known to be normal (answer b).

2) C. H0: mu = 519, H_1: mu not equal to 519, reject if z> 1.96 or z< -1.96.

3) Yes. There is enough evidence to support the claim that the true mean is not 519 ml.

Step-by-step explanation:

1) When the population follows a normal distribution, it is correct to assume a normal distribution for the sample mean.

2) As it is a two-tailed decision rule, we are interested in detecting a significant difference below and above the mean. This is why we use the unequal sign in the alternative hypothesis.

The null hypothesis state that there is not significant difference from 519.

The critical value for a significance level of 5% is z=1.96.

$H_0: \mu=519\\\\H_a:\mu\neq 519$

3) The claim is that the true mean is not 519 ml.

Then, the null and alternative hypothesis are:

$H_0: \mu=519\\\\H_a:\mu\neq 519$

The significance level is 0.05.

The sample has a size n=16.

The sample mean is M=522.

The standard deviation of the population is known and has a value of σ=6.

We can calculate the standard error as:

$\sigma_M=\dfrac{\sigma}{\sqrt{n}}=\dfrac{6}{\sqrt{16}}=1.5$

Then, we can calculate the z-statistic as:

$z=\dfrac{M-\mu}{\sigma_M}=\dfrac{522-519}{1.5}=\dfrac{3}{1.5}=2$

This test is a two-tailed test, so the P-value for this test is calculated as:

$\text{P-value}=2\cdot P(z>2)=0.046$

As the P-value (0.046) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the true mean is not 519 ml.

7 0

3 years ago