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navik [9.2K]
3 years ago
7

Evaluate what research has shown about the relationship between intelligence, biological sex, and race or ethnicity. Use evidenc

e presented in the text to support your claim. Be sure to also discuss what psychologists believe.​
Advanced Placement (AP)
2 answers:
Ksenya-84 [330]3 years ago
8 0

Answer:

Research has shown that there are both similarities and differences between genders and race or ethnicity when it comes to intelligences. There are many gender differences in intelligence some categories are more established by women and others by men. Woman score higher in spelling, verbal ability, emotion-detecting ability, and nonverbal memory. Men score higher in math and spatial aptitude. Research demonstrates that ethnic groups differ overall in their average intelligence score. They vary because of stereotypes, heritability, cultural bias, and socioeconomic disadvantages.

Explanation:

Amiraneli [1.4K]3 years ago
7 0

Answer:

In the search for a better understanding of genetic and environmental interactions as determinants of health, certain fundamental aspects of human identity pose both a challenge and an opportunity for clarification. Sex/gender and race/ethnicity are complex traits that are particularly useful and important because each includes the social dimensions necessary for understanding its impact on health and each has genetic underpinnings, to varying degrees.

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2 years ago
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Hello, I need help with a calculus FRQ. My teacher has given a hint that this last part has to do with the intermediate value th
lesya [120]

Answer:

Yes, at a time t such that (√2)/2 ≤ t ≤ 2.

Explanation:

To answer the question

Therefore, where the domain of the function is the set of all real numbers x for which f(x) is a real number we have

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C(t) = t\times e^{4-t^2} \ for \ 0\leq t\leq 2

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0\times e^{4-0^2} = 0 and 2\times e^{4-2^2} = 2

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\frac{\mathrm{d} \left (t\times e^{4-t^2}   \right )}{\mathrm{d} x}=0

From which we have;

\frac{\mathrm{} e^{4-t^2} - t\times e^{4-t^2} \times2\times t }{(e^{4-t^2} )^2}=0

e^{4-t^2} - t\times e^{4-t^2} \times2\times t }=0

e^{4-t^2}(1 - t\times2\times t })=0\\e^{4-t^2}(1 - 2\times t^2 })=0\\

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For Brandon

For 0 ≤ t ≤ 1, 1 ≤ B(t) ≤8 and for 1 < t ≤ 2, 8 < B(t) ≤ 1.5

1 ≤ f(x) ≤ 1.5

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