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3 years ago

16x+9=9y-2x solve for y

1 answer:

7 0

Ok, so for this problem you want to get Y all by itself.
First you’ll want to move your X variables to one side. You do that by adding 2X to both sides of the equation. This will cancel out the -2X and it will turn the 16X into an 18X.
You’re equation is now,
18X+9=9Y
Next you’ll want to divide both sides by 9 so that you’ll get Y all by itself.
Your equation should look like this.
Y=2X+1!
And that’s your answer:) I hope this helped!

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3 years ago

Un escalador sube 225m de altura. El asenso lo hizo en 3 etapas. En la primera subio un quinto en la, en la tercera un cuarto. ¿

kiruha [24]

Answer:

Segunda etapa= 123.75 metros

Step-by-step explanation:

Altura total= 225 metros

En la primera estapa subió el 20% (un quinto):

Primera etapa= 225*0.2= 45 metros

En la tercera etapa subió 25% (un cuarto):

Tercera etapa= 225*0.250 56.25 metros

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3 years ago

A simple random sample of 36 men from a normally distributed population results in a standard deviation of 10.1 beats per minute

GREYUIT [131]

Answer:

(a) Null Hypothesis, $H_0$ : $\sigma$ = 10 beats per minute

Alternate Hypothesis, $H_A$ : $\sigma\neq$ 10 beats per minute

(b) The value of chi-square test statistics is 35.704.

(c) P-value = 0.4360.

(d) We conclude that the pulse rates of men have a standard deviation equal to 10 beats per minute.

Step-by-step explanation:

We are given that a simple random sample of 36 men from a normally distributed population results in a standard deviation of 10.1 beats per minute.

If the range rule of thumb is applied to that normal range, the result is a standard deviation of 10 beats per minute.

Let $\sigma$ = population standard deviation for the pulse rates of men.

(a) So, Null Hypothesis, $H_0$ : $\sigma$ = 10 beats per minute {means that the pulse rates of men have a standard deviation equal to 10 beats per minute}

Alternate Hypothesis, $H_A$ : $\sigma\neq$ 10 beats per minute {means that the pulse rates of men have a standard deviation different from 10 beats per minute}

The test statistics that will be used here is One-sample chi-square test for standard deviation;

T.S. = $\frac{(n-1)\times s^{2} }{\sigma^{2} }$ ~ $\chi^{2}__n_-_1$

where, s = sample standard deviation = 10.1 beats per minute

n = sample of men = 36

So, the test statistics = $\frac{(36-1)\times 10.1^{2} }{10^{2} }$ ~ $\chi^{2}__3_5$

= 35.70 4

(b) The value of chi-square test statistics is 35.704.

(c) Also, the P-value of the test statistics is given by;

P-value = P( $\chi^{2}__3_5$ > 35.704) = 0.4360

(d) Since the P-value of our test statistics is more than the level of significance as 0.4360 > 0.10, so we have insufficient evidence to reject our null hypothesis as the test statistics will not fall in the rejection region.

Therefore, we conclude that the pulse rates of men have a standard deviation equal to 10 beats per minute.

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4 years ago