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maria [59]
2 years ago
15

Is Is y=2x+7 proportional?

Mathematics
1 answer:
inn [45]2 years ago
7 0

Answer:

no

Step-by-step explanation:

Proportional must go through (0,0)

0 = 2(0) +7

0 = 7

This is not true so this is not proportional

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Compré 5 distintos tipos de semillas de flores para sembrarlas en el jardin una tras de otra.
Colt1911 [192]

Responder:

120 maneras.

Explicacion paso a paso:

Cinco tipos diferentes de semillas.

El numero de formas de organizar, n, a diferencia de los objetos en una linea es n!

Entonces, para sembrar 5 tipos diferentes de flores, serian 5!

5! = 5 * 4 * 3 * 2 * 1 = 120 vias.

Por lo tanto, hay 120 formas de sembrar las cinco semillas.

Hablo ingles y escribo ingles. Espero que entiendas. Acabo de utilizar el traductor de Google para intentar que la respuesta sea mas clara para ti :).

Por favor, califique con 5 estrellas, gracias, y deme lo mejor (Brainliest). Gracias.

8 0
3 years ago
Plzzzz helppppppppp and free kodak
11Alexandr11 [23.1K]

Answer:

77.77

Step-by-step explanation:

you subtract 16 (prediction) - 9 (actual) and divide it by the actual.

7 0
3 years ago
Read 2 more answers
How to solve y for this problem:
True [87]
16x+9=9y-2x\ \ |Add\ 2x\ to\ both\ sides\\\\
16x+9+2x=9y-2x+2x\\\\
18x+9=9y\ \ \ |Divide\ by\ 9\\\\
\frac{18x}{9}+\frac{9}{9}=\frac{9y}{9}\\\\
2x+1=y\\\\
Solution\ is\ y=2x+1.
6 0
2 years ago
Statistics In the manual “How to Have a Number One the Easy Way,” it is stated that a song “must be no longer than three minutes
dimaraw [331]

Answer:

We conclude that the sample is from a population of songs with a mean greater than 210 seconds.

Step-by-step explanation:

We are given that a simple random sample of 40 current hit songs results in a mean length of 252.5 seconds.

Assume that the standard deviation of song lengths is 54.5 sec.

Let \mu = <u><em>population mean length of the songs</em></u>

So, Null Hypothesis, H_0 : \mu \leq 210 seconds      {means that the sample is from a population of songs with a mean smaller than or equal to 210 seconds}

Alternate Hypothesis, H_A : \mu > 210 seconds      {means that the sample is from a population of songs with a mean greater than 210 seconds}

The test statistics that will be used here is <u>One-sample z-test statistics </u>because we know about population standard deviation;

                              T.S.  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean length of songs = 252.5 seconds

            \sigma = population standard deviation = 54.5 seconds

            n = sample of current hit songs = 40

So, <u><em>the test statistics</em></u> =  \frac{252.5-210}{\frac{54.5}{\sqrt{40} } }

                                   =  4.932

The value of z-test statistics is 4.932.

Now, at 0.05 level of significance, the z table gives a critical value of 1.645 for the right-tailed test.

Since the value of our test statistics is more than the critical value of z as 4.932 > 1.645, so <u><em>we have sufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.

Therefore, we conclude that the sample is from a population of songs with a mean greater than 210 seconds.

4 0
2 years ago
Please help on the production one.
kirza4 [7]
5x + 8000 = 3x + 10000
x = 1000
5 0
3 years ago
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