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nekit [7.7K]

Answer:

Yes it is absolutely right.

3 0

3 years ago

Aliyah spent half of her weekly allowance playing arcade games. To earn more money her parents let her mow the lawn for $7. What

Fudgin [204]

Answer:

who mows the lawn for 7 dollars :/

Step-by-step explanation:

4 0

3 years ago

Hello can someone please help?

Ray Of Light [21]

The answer is D
Hope this helps

7 0

3 years ago

The local swim team is considering offering a new semi-private class aimed at entry-level swimmers, but needs a minimum number o

SCORPION-xisa [38]

Answer:

The significance level is $\alpha=0.01$ and since we are conducting a right tailed test we need to find a critical value who accumulate 0.01 of the area in the right of the normal standard distribution and we got:

$z_{\alpha/2}= 2.326$

So we reject the null hypothesis is $z>2.326$

Step-by-step explanation:

For this case we define the random variable X as the number of entry-level swimmers and we are interested about the true population mean for this variable . On specific we want to test this:

Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu > 15$

And the statistic is given by:

$z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

The significance level is $\alpha=0.01$ and since we are conducting a right tailed test we need to find a critical value who accumulate 0.01 of the area in the right of the normal standard distribution and we got:

$z_{\alpha/2}= 2.326$

So we reject the null hypothesis is $z>2.326$

7 0

3 years ago

3x+10y=13<br> 4(-2y+x)-9x=13

mart [117]

Answer:

$x=-9$