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3 years ago

What were some provisions of the Dred Scott case? Select three responses

Mathematics

2 answers:

aleksley [76]3 years ago

7 0

Answer:

Following were the provisions of Dred Scott case:

No black person, free or enslaved, could be a US citizen.
There could be no legal ban on slavery on the territories.
The Missouri Compromise was unconstitutional.

Step-by-step explanation:

Dred scott was the slave of an army surgeon. He accompanied his owner to Illinois, which had become a free state at that time. However, he later returned back to Missouri, which was a slave state. In 1846, Dred Scott sued for their freedom in a St. Louis court. According to him, as he had been living in a free state for some time, he should be freed from the slavery.

After 11 years, Dred Scott's case eventually reached the US Supreme court, which ruled against him.

Pavlova-9 [17]3 years ago

4 0

Answer:

A,C,E

Step-by-step explanation:

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Trinity collected 2,400 grams of pennies in a fundraiser. Each penny has a mass of 2.5 grams. How much money did Trinity raise?

sp2606 [1]

Answer:

960

Step-by-step explanation:

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3 years ago

A 26' long painting is how many yards long?

mamaluj [8]

Answer:

The correct answer is 8 2/3yards

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3 years ago

Read 2 more answers

The Blue Diamond Company advertises that their nut mix contains (by weight) 40% cashews, 15% Brazil nuts, 20% almonds and only 2

jek_recluse [69]

Answer:

At the 0.05 level of significance, the claim made by Blue Diamond is true

Step-by-step explanation:

Given that the Blue Diamond Company advertises that their nut mix contains (by weight) 40% cashews, 15% Brazil nuts, 20% almonds and only 25% peanuts.

$H_0: The mix are as per percents given\\H_a:Atleast two differ from the above percent.$

(Two tailed chi square test at 5% significance level)

Chi square is calculated as (Obs_exp)^2/Expected

Cashews Brazil nuts Almonds Peanuts Total

Observed 5 6 6 3 20

Expected 40% 15% 20% 25% 1

Expected 8 3 4 5 20

Chi square 1.125 3 1 0.8 5.925

df = 3

p value = 0.1153

Since p value >0.0 we accept null hypothesis.

4 0

3 years ago

What is the inverse of f(x)=3x+6 ?

scZoUnD [109]

Answer:

$\boxed{\sf \ \ f^{-1}(x)=\dfrac{x-6}{3} \ \ }$

Step-by-step explanation:

hello,

we can write

$fof^{-1}(x)=x \ and \ fof^{-1}(x)=f(f^{-1}(x))=3f^{-1}(x)+6 \ so\\3f^{-1}(x)+6=x \ \ subtract \ 6\\3f^{-1}(x)=x-6 \ \ divide \ \ by \ \ 3\\ f^{-1}(x)=\dfrac{x-6}{3}$

hope this helps

3 0

3 years ago

The time for this event for boys in secondary school is known to possess a normal distribution with a mean of 460 seconds and a

AlekseyPX

Answer:

The time that the boys need to beat in order to earn a certificate of recognition from the fitness association is 511.264 seconds.

Step-by-step explanation:

We are given that the time for this event for boys in secondary school is known to possess a normal distribution with a mean of 460 seconds and a standard deviation of 40 seconds.

The fitness association wants to recognize the fastest 10% of the boys with certificates of recognition.

Let X = time for this event for boys in secondary school

SO, X ~ Normal( $\mu=460,\sigma^{2} =40^{2}$ )

The z-score probability distribution for normal distribution is given by;

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

where, $\mu$ = mean time = 460 seconds

$\sigma$ = standard deviation = 40 seconds

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, it is given that the fitness association wants to recognize the fastest 10% of the boys with certificates of recognition, which means;

P(X > x) = 0.10 {where x is the required time which boy need to beat}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-460}{40}$ ) = 0.10