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2 years ago

I don't know just someone answer this for me thanks

Mathematics

2 answers:

lesya692 [45]2 years ago

8 0

Answer:

the answer would be 43 degrees.

Step-by-step explanation:

Tanzania [10]2 years ago

6 0

The answer is 43 degrees

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consider the following scenario: Assume daily protein intake values in a population of athletes are known to have a standard dev

Kipish [7]

Answer:

The 95% confidence interval for the population mean daily protein intake is between 69.97g and 84.03g.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{58.6}{\sqrt{267}} = 7.03$

The lower end of the interval is the sample mean subtracted by M. So it is 77 - 7.03 = 69.97g.

The upper end of the interval is the sample mean added to M. So it is 77 + 7.03 = 84.03g.

The 95% confidence interval for the population mean daily protein intake is between 69.97g and 84.03g.

5 0

3 years ago

Need help ASAP PLEASE DUE SOON!!!

beks73 [17]

Answer:

Step-by-step explanation:

all u need to do is to cube root 1331, because it is a cube. hope i helped.

7 0

2 years ago

The area of a rectangular invitation card is 3/10 square foot. Given that the length of the card is 4/5 foot find its perimeter

Andrei [34K]

The perimeter would be 3/8+3/8+4/5+4/5= 2.35 or 2 7/20 square foot.
I hope this helps.
YOU'RE WELCOME :D

8 0

2 years ago

Prove the following DeMorgan's laws: if LaTeX: XX, LaTeX: AA and LaTeX: BB are sets and LaTeX: \{A_i: i\in I\} {Ai:i∈I} is a fam

MariettaO [177]

$X-(A\cup B)=(X-A)\cap(X-B)$

I'll assume the usual definition of set difference, $X-A=\{x\in X,x\not\in A\}$ .

Let $x\in X-(A\cup B)$ . Then $x\in X$ and $x\not\in(A\cup B)$ . If $x\not\in(A\cup B)$ , then $x\not\in A$ and $x\not\in B$ . This means $x\in X,x\not\in A$ and $x\in X,x\not\in B$ , so it follows that $x\in(X-A)\cap(X-B)$ . Hence $X-(A\cup B)\subset(X-A)\cap(X-B)$ .

Now let $x\in(X-A)\cap(X-B)$ . Then $x\in X-A$ and $x\in X-B$ . By definition of set difference, $x\in X,x\not\in A$ and $x\in X,x\not\in B$ . Since $x\not A,x\not\in B$ , we have $x\not\in(A\cup B)$ , and so $x\in X-(A\cup B)$ . Hence $(X-A)\cap(X-B)\subset X-(A\cup B)$ .

The two sets are subsets of one another, so they must be equal.

$X-\left(\bigcup\limits_{i\in I}A_i\right)=\bigcap\limits_{i\in I}(X-A_i)$

The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices $i\in I$ .

Proof of one direction for example:

Let $x\in X-\left(\bigcup\limits_{i\in I}A_i\right)$ . Then $x\in X$ and $x\not\in\bigcup\limits_{i\in I}A_i$ , which in turn means $x\not\in A_i$ for all $i\in I$ . This means $x\in X,x\not\in A_{i_1}$ , and $x\in X,x\not\in A_{i_2}$ , and so on, where $\{i_1,i_2,\ldots\}\subset I$ , for all $i\in I$ . This means $x\in X-A_{i_1}$ , and $x\in X-A_{i_2}$ , and so on, so $x\in\bigcap\limits_{i\in I}(X-A_i)$ . Hence $X-\left(\bigcup\limits_{i\in I}A_i\right)\subset\bigcap\limits_{i\in I}(X-A_i)$ .

4 0

2 years ago

HELPP FAST WILL GIVE BRAINLIEST ANSWER AND WRITE SOMETHING NICE ON PROFILE!!

Temka [501]

#3)answer is choice D.77+W<_95;W<_18
#4)answer is choice A. t<7

7 0

3 years ago

Read 2 more answers