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2 years ago

What is the solution to the system of equation?

Mathematics

1 answer:

dusya [7]2 years ago

4 0

Answer:

(0, -4)

Step-by-step explanation:

The solution to the system is the point that is on both lines. It is where the lines intersect (cross). They intersect at the point (0, -4)

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Which of the following uses logical deduction to prove the truth (or falsehood) of a specific statement?

ser-zykov [4K]

Direct proof. A direct proof is a way of establishing the truth or falsehood of a given mathematical statement by a combination of established facts, usually axioms, lemmas and theorems. It assumes the hypothesis of a conjecture is true and then uses a series of logical deductions to prove that the conclusion as well.

7 0

4 years ago

Kiersten runs a website that helps people learn programming. Every month, Kiersten receives a subscription fee of \$10$10dollar

NemiM [27]

700009000720067261836

5 0

3 years ago

Read 2 more answers

I arrive at a bus stop at a time that is normally distributed with mean 08:00 and SD 2 minutes. My bus arrives at the stop at an

Nimfa-mama [501]

Answer:

0.0485 = 4.85% probability that you miss the bus.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

When two normal distributions are subtracted, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

In this question:

We have to find the distribution for the difference in times between when you arrive and when the bus arrives.

You arrive at 8, so we consider the mean 0. The bus arrives at 8:05, 5 minutes later, so we consider mean 5. This means that the mean is:

$\mu = 0 - 5 = -5$

The standard deviation of your arrival time is of 2 minutes, while for the bus it is 3. So

$\sigma = \sqrt{2^2 + 3^2} = \sqrt{13}$

The bus remains at the stop for 1 minute and then leaves. What is the chance that I miss the bus?

You will miss the bus if the difference is larger than 1. So this probability is 1 subtracted by the pvalue of Z when X = 1.

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1 - (-5)}{\sqrt{13}}$

$Z = \frac{6}{\sqrt{13}}$

$Z = 1.66$

$Z = 1.66$ has a pvalue of 0.9515

1 - 0.9515 = 0.0485

0.0485 = 4.85% probability that you miss the bus.

5 0

3 years ago

What is the value of the expression (9+15)/3+2?

KIM [24]

(9+15)/3+2
24/3+2
8+2
10

6 0

3 years ago

Read 2 more answers

A coffee packaging plant claims that the mean weight of coffee in its containers is at least 32 ounces. A random sample of 15 co

Luda [366]

Answer:

We conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.

Step-by-step explanation:

We are given that a coffee packaging plant claims that the mean weight of coffee in its containers is at least 32 ounces.

A random sample of 15 containers were weighed and the mean weight was 31.8 ounces with a sample standard deviation of 0.48 ounces.

Let $\mu$ = mean weight of coffee in its containers.

SO, Null Hypothesis, $H_0$ : $\mu \geq$ 32 ounces {means that the mean weight of coffee in its containers is at least 32 ounces}

Alternate Hypothesis, $H_A$ : $\mu$ < 32 ounces {means that the mean weight of coffee in its containers is less than 32 ounces}

The test statistics that would be used here One-sample t-test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean weight = 31.8 ounces

s = sample standard deviation = 0.48 ounces

n = sample of containers = 15

So, the test statistics = $\frac{31.8 -32}{\frac{0.48}{\sqrt{15} } }$ ~ $t_1_4$

= -1.614

The value of t test statistics is -1.614.

Now, at 0.01 significance level the t table gives critical value of -2.624 at 14 degree of freedom for left-tailed test.

Since our test statistic is more than the critical value of t as -1.614 > -2.624, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the mean weight of coffee in its containers is at least 32 ounces which means that the data support the claim.

8 0

4 years ago