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3 years ago

Please help and if u want to do this ixl msg me

Mathematics

1 answer:

Illusion [34]3 years ago

7 0

Answer:

I need this to

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Splitting Squares (square root)<br><br> Number 34 (picture above)

miskamm [114]

Answer: $3rp\sqrt{5}$

<u>Step-by-setp explanation:</u>

$\sqrt{45r^{2}p{2}}$

= $\sqrt{3*3*5*r*r*p*p}$

= $3*r*p\sqrt{5}$

= $3rp\sqrt{5}$

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3 years ago

Triangle ABC was transformed to create triangle A'B'C'.

makkiz [27]

Answer:

The answer is F(x, y-7).

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3 years ago

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.Martin is an after-school math tutor. He noticed that 6 people still needed help and only Two-fifths of the tutoring session ti

Ugo [173]

Answer:

6/15 divided by 6

Step-by-step explanation:

$\frac{2}{5} / 6\\= \frac{6}{15} / 6\\$

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3 years ago

Find the distance and midpoint between (3,6) and (5,11)

zepelin [54]

Answer:

((3+5)/2,(6+11)/2)=(4,8.5)

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3 years ago

A study of pollutants showed that certain industrial emissions should not exceed 2.3 parts per million. You believe a particular

Musya8 [376]

Answer:

$t=\frac{3.3-2.3}{\frac{0.4}{\sqrt{9}}}=7.5$

The degrees of freedom are given by:

$df=n-1=9-1=8$

And the p value would be:

$p_v =P(t_{(8)}>7.5)=0.0000346$

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 2.3 ppm the safe limit

Step-by-step explanation:

Information provided

$\bar X=3.3$ represent the sample mean in ppm

$s=0.4$ represent the sample standard deviation

$n=9$ sample size

$\mu_o =2.3$ represent the value to verify

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

System of hypothesis

We want to verify if the true mean is higher than 2,3 ppm, the system of hypothesis would be:

Null hypothesis: $\mu \leq 2.3$

Alternative hypothesis: $\mu > 2.3$

The statistic for this case is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$t=\frac{3.3-2.3}{\frac{0.4}{\sqrt{9}}}=7.5$

The degrees of freedom are given by:

$df=n-1=9-1=8$

And the p value would be:

$p_v =P(t_{(8)}>7.5)=0.0000346$

Since the p value is lower than the significance level we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 2.3 ppm the safe limit

6 0

4 years ago