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3 years ago

Plzzz help???????????

Mathematics

2 answers:

uysha [10]3 years ago

6 0

Answer:

Its either C. or D.

Step-by-step explanation:

Lina20 [59]3 years ago

5 0

Answer:

the answer is B................

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PLZ HELP ASAP, LINKS AND ABSURD ANSWERS WILL BE REPORTED! I WILL MARK BRAINLIEST!!

Stels [109]

The closest answer I can see you type is: 35/1.66666666

This is because you divide 21 by 18 which (if you divide 21 by 1.666666 you get 18) and whatever you do to 1 side, you do to the other, then you get 35/1.6666666

7 0

3 years ago

In the inequality, what are all the possible values of x?

Marat540 [252]

Answer:

B) $\displaystyle x ≤ 0$

Step-by-step explanation:

2(x + 12) − 16 ≤ 8

$\displaystyle \frac{2(x + 12)}{2} ≤ \frac{24}{2} \\ \\ x + 12 ≤ 12 \\ \\ x ≤ 0$

I am joyous to assist you anytime.

7 0

3 years ago

A student bought 34 pencils for school.if he sharpened 16 of the pencils before school whats is his ratio of unsharpened pencils

Sergio [31]

he bought 34 pencils which I suppose aren't already sharpened

so if he sharpens 16 of them just subtract 16 from 34 (34-16) which equals 18

so the ratio would be 16:18

but if u need to simplify it would be 8:9

3 0

3 years ago

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

Given the function h(w)=3w2−11w, solve for h(w)=−6.

Vesnalui [34]

Answer:

w = 2/3 AND w = 3

Step-by-step explanation:

Set 3w2-11w = -6

So 3w2-11w+6 = 0

Now factor: (3w-2)(w-3) = 0

So 3w-2 = 0 AND w-3 = 0

So w=2/3 AND w = 3

8 0

3 years ago