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fgiga [73]
3 years ago
15

Based on the graph of the following system, determine the solution​

Mathematics
1 answer:
Assoli18 [71]3 years ago
3 0

Answer:

(2,4)

Step-by-step explanation:

The solution to graphed systems is quite easy to find, as the solution is the intersection between the lines. The intersection in this graph is (2,4).

Hope this helps!

You might be interested in
( − 2 7 ) ( 5 − 8 ) Please solve .
stepan [7]

Answer:

81

Step-by-step explanation:

We have to solve the following multiplication: (−27) (5−8)

First, we are going to simplify the second parenthesis, before performing the multiplication:

(−27) (5−8) = (−27)(-3) = 81.

The result is positive given that minus times minus equals plus.

6 0
4 years ago
Read 2 more answers
Test the claim that the mean GPA of night students is larger than 2 at the .025 significance level. The null and alternative hyp
exis [7]

Answer:

H_0: \, \mu = 2.

H_1:\, \mu > 2.

Test statistics: z \approx 2.582.

Critical value: z_{1 - 0.025} \approx 1.960.

Conclusion: reject the null hypothesis.

Step-by-step explanation:

The claim is that the mean \mu is greater than 2. This claim should be reflected in the alternative hypothesis:

H_1:\, \mu > 2.

The corresponding null hypothesis would be:

H_0:\, \mu = 2.

In this setup, the null hypothesis H_0:\, \mu = 2 suggests that \mu_0 = 2 should be the true population mean of GPA.

However, the alternative hypothesis H_1:\, \mu > 2 does not agree; this hypothesis suggests that the real population mean should be greater than \mu_0= 2.

One way to test this pair of hypotheses is to sample the population. Assume that the population mean is indeed \mu_0 = 2 (i.e., the null hypothesis is true.) How likely would the sample (sample mean \overline{X} = 2.02 with sample standard deviation s = 0.06) be observed in this hypothetical population?

Let \sigma denote the population standard deviation.

Given the large sample size n = 60, the population standard deviation should be approximately equal to that of the sample:

\sigma \approx s = 0.06.

Also because of the large sample size, the central limit theorem implies that Z= \displaystyle \frac{\overline{X} - \mu_0}{\sigma / \sqrt{n}} should be close to a standard normal random variable. Use a Z-test.

Given the observation of \overline{X} = 2.02 with sample standard deviation s = 0.06:

\begin{aligned}z_\text{observed}&= \frac{\overline{X} - \mu_0}{\sigma / \sqrt{n}} \\ &\approx \frac{\overline{X} - \mu_0}{s / \sqrt{n}} = \frac{2.02 - 2}{0.06 / \sqrt{60}} \approx 2.582\end{aligned}.

Because the alternative hypothesis suggests that the population mean is greater than \mu_0 = 2, the null hypothesis should be rejected only if the sample mean is too big- not too small. Apply a one-sided right-tailed z-test. The question requested a significant level of 0.025. Therefore, the critical value z_{1 - 0.025} should ensure that P( Z > z_{1 - 0.025}) = 0.025.

Look up an inverse Z table. The z_{1 - 0.025} that meets this requirement is z_{1 - 0.025} \approx 1.960.

The z-value observed from the sample is z_\text{observed}\approx 2.582, which is greater than the critical value. In other words, the deviation of the sample from the mean in the null hypothesis is sufficient large, such that the null hypothesis needs to be rejected at this 0.025 confidence level in favor of the alternative hypothesis.

3 0
3 years ago
What is the down payment total for a $189,700 home at 15% down
svet-max [94.6K]

Answer:

$28,455

Step-by-step explanation:

Change 15% into a decimal.

(15)(1/100)=0.15

***(1/100) because 15 part of the whole 100 percent/.

(189,700)(0.15)

=28455

4 0
4 years ago
Read 2 more answers
En la inaguracion de unos juegos olimpicos, los 38 representantes de una nacion quieren desfilar formando columnas de 6¿es posib
worty [1.4K]

Answer:

a) No

b) Sí

Step-by-step explanation:

a) En la inauguración de unos Juegos Olímpicos, los 38 representantes de una nación quieren desfilar en columnas de 6, ¿es posible?

Esto se calcula como

38 representantes / 6 columnas

= 6 2/6 = 6 1/3 columnas

No, no es posible, tendríamos 7 columnas con 6 columnas con representantes iguales de 6 y la séptima columna tendría solo 2 representantes allí

b) ¿Cómo podrían desfilar en columnas iguales?

Sí pueden

Esto se calcula como 38 representantes ÷ 2

= 19 columnas

Podemos tener 19 columnas con 2 representantes cada una

4 0
3 years ago
If theta is a positive acute angles find theta when <br>2 sin^2 theta + 3 cos theta =3​
MArishka [77]

Answer:

Theta = 0, 60 degrees.

Step-by-step explanation:

2 sin^2 theta + 3 cos theta = 3​

Using the identity sin ^2 theta = 1 - cos^2 theta :

2(1 - cos^2 theta) + 3 cos theta  = 3

2 - 2cos^2 theta) + 3 cos theta  = 3

2 cos^2 theta - 3 cos theta   + 3 - 2 = 0

2 cos^2 theta - 3 cos theta   + 1 = 0

(2 cos theta - 1)(cos theta - 1) = 0

cos theta = 1/2 , cos theta = 1

Theta = 60 degrees or theta = 0

6 0
3 years ago
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