Need help on geometry

Mathematics

2 answers:

Rama09 [41]3 years ago

7 0

ASA, SAS, AAS,

if LA stands for leg-angle then you could do that too

goldfiish [28.3K]3 years ago

4 0

I believe B-E as for the other, not sure

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Children of mothers who smoked during pregnancy scored nine points lower on intelligence tests at ages three and four than child

eduard

Answer:

All of the options apply.

Step-by-step explanation:

In the example given in the question, it is stated that mothers' smoking during pregnancy can affect childs' IQ and can cause lower scores on test.

All of the given options can be given as evidence for the link between smoking and lower IQ. The diet especially throughout childhood has significant effect on the development of the brain and the prefrontal cortex.

The psychological condition that the child is in, the environment in the house that he/she grows up in affects the IQ and the EQ.

The IQ can be affected by outside factors but it also has a relation with the genes that the child takes from his/her parents.

The socioeconomic status of the mother can affect the environment that the child grows up in, the people he/she interacts with and the opportunities that he/she gets to stimulate the brain in different ways to encourage growth.

I hope this answer helps.

8 0

3 years ago

A researcher was interested in seeing how many names a class of 38 students could remember after playing a name game After playi

Andreas93 [3]

Answer:

Proportion of the students recalled more than 15 names is 91.77%.

Step-by-step explanation:

We are given that a researcher was interested in seeing how many names a class of 38 students could remember after playing a name game After playing the name game, the students were asked to recall as many first names of fellow students as possible.

The mean number of names recalled was 19.41 with a standard deviation of 3.17.

Let X = number of names recalled

SO, X ~ N( $\mu = 19.41,\sigma^{2} = 3.17^{2}$ )

The z-score probability distribution is given by ;

Z = $\frac{X-\mu}{\sigma} } }$ ~ N(0,1)

where, $\mu$ = mean number of names recalled = 19.41

$\sigma$ = standard deviation = 3.17

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, proportion of the students recalled more than 15 names is given by = P(X > 15 names)

P(X > 15) = P( $\frac{X-\mu}{{\sigma} } }$ > $\frac{15-19.41}{3.17} }$ ) = P(Z > -1.39)