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3 years ago

What is the variable "x" in the equation, 7/2=x/2, and why?

1 answer:

7 0

X is equal to 7. Since you are trying to make them equal to each other, the denominator is already the same, so you just need to make the numerator the same too.

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I need help on my math homework

xxTIMURxx [149]

Answer: the first one is True and the second one is false

Step-by-step explanation:

3 + 5x = 23

so replace x with 4 you get 3 + 5 * 4 = 3 + 20 = 23

this is equal to 23 so it is true

2x - 6 = 9

do the same you get

2 * 8 - 6 = 9

16 - 6 = 10

the answer is 10 not nine therefore it is false

7 0

3 years ago

Find the value of y that makes each equation true for all values of x.

Soloha48 [4]

Okay I'm a little lost because of the . next to the 2 but if they are separate your answers would be

3x+6x=yx (x=0)

c+5+11=12+7 (x=x=−y+c+5+<span>
</span>

3 0

3 years ago

Write a word problem for which you would use the equation 2/5x + 6 = 56

velikii [3]

Answer:

Isha is at an arcade. There is a base fee of $6 to enter the game room. Every game costs $0.4 or 40 cents to play. If Isha ends up spending $56 at the arcade, how many games did she play?

An explanation:

The base fee of $6 represents the constant 6 in the equation. Every game costs $0.4, which equals 2/5.

Hope this helps! If you don't like this one I might be able to think of another one :)

7 0

3 years ago

SOMEONE PLZ HELP?????

Art [367]

(43)3 multiplied by x=y

4 0

3 years ago

Read 2 more answers

Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of t

Likurg_2 [28]

Answer:

This data suggest that there is more variability in low-dose weight gains than in control weight gains.

Step-by-step explanation:

Let $\sigma_{1}^{2}$ be the variance for the population of weight gains for rats given a low dose, and $\sigma_{2}^{2}$ the variance for the population of weight gains for control rats whose diet did not include the insecticide.

We want to test $H_{0}: \sigma_{1}^{2} = \sigma_{2}^{2}$ vs $H_{1}: \sigma_{1}^{2} > \sigma_{2}^{2}$ . We have that the sample standard deviation for $n_{2} = 22$ female control rats was $s_{2} = 28$ g and for $n_{1} = 18$ female low-dose rats was $s_{1} = 51$ g. So, we have observed the value

$F = \frac{s_{1}^{2}}{s_{2}^{2}} = \frac{(51)^{2}}{(28)^{2}} = 3.3176$ which comes from a F distribution with $n_{1} - 1 = 18 - 1 = 17$ degrees of freedom (numerator) and $n_{2} - 1 = 22 - 1 = 21$ degrees of freedom (denominator).

As we want carry out a test of hypothesis at the significance level of 0.05, we should find the 95th quantile of the F distribution with 17 and 21 degrees of freedom, this value is 2.1389. The rejection region is given by {F > 2.1389}, because the observed value is 3.3176 > 2.1389, we reject the null hypothesis. So, this data suggest that there is more variability in low-dose weight gains than in control weight gains.

8 0

3 years ago