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gregori [183]
3 years ago
15

I WOULD APPRECIATE IT A LOT IF SOMEONE COULD ANSWER THESE PROBLEMS!!! THANK YOU IF YOU CAN!

Mathematics
1 answer:
romanna [79]3 years ago
8 0
10. C.

11. D.

12. C.

13. C.

14. A.

15. D.
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For the statement​ below, write the claim as a mathematical statement. State the null and alternative hypotheses and identify wh
vfiekz [6]

The question is not complete, so i have attached it.

Answer:

A) Option B: μ ≤ 19,000

B) Option C:

H0: μ ≤ 19,000

Ha: μ > 19,000

C) Option D: The null hypothesis H0: μ ≤ 19,000 is the claim

Step-by-step explanation:

A) From the question, we are told that an amusement park claims that the mean daily attendence at the park is at most 19,000 people.

Now, since it's at most 19000 people, it means that they are either 19000 or less. So the claim is;

μ ≤ 19,000 which is option B

B) Now, the claim is usually the null hypothesis. Thus, the null hypothesis is;

H0: μ ≤ 19,000

While the alternative hypothesis which states the opposite of the claim is;

Ha: μ > 19,000

So, option C is the correct answer

C) The claim is μ ≤ 19,000

Now, this also happens to be the null hypothesis.

Thus, the null hypothesis H0: μ ≤ 19,000 is the claim.

So option D is correct.

6 0
4 years ago
a photo album contains small and large photographs. Each large photograph has side lengths that are twice the side lengths of ea
zysi [14]
Take the area of the small photograph and divide it by four, assuming it's a square, each side would then be 6 inches. The large photograph has double the side length of the small one, so 6 multiplied by 2 is 12. To find the area of the large (square) photograph, multiply 12 by 12 as two of the sides, and the area of the large photo is 144 square inches
4 0
3 years ago
Read 2 more answers
What is the midpoint coordinates of segment HX H(13,8) X(-6,-6)
Setler79 [48]
Midpoint of (x1,y1) and (x2,y2) is

(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

so the midpoint of HX is the mdpoint of (13,8) and (-6,-6)
which is

(\frac{13-6}{2},\frac{8-6}{2}=
(\frac{7}{2},\frac{2}{2}=
(3.5,1)

the midpoint is (3.5,1)
7 0
4 years ago
Plzzz help plzzzzzzzzzzzzzz
mezya [45]

Given:

The figure of two quadrilaterals.

In ABCD,AB=18,BC=20,CD=22,AD=24

In EFGH,EF=27,FG=30,GH=34, EH=36

To find:

Whether the figures are congruent, similar or neither.

Solution:

Ratio of corresponding sides are:

\dfrac{AB}{EF}=\dfrac{18}{27}

\dfrac{AB}{EF}=\dfrac{2}{3}

Similarly,

\dfrac{BC}{FG}=\dfrac{20}{30}

\dfrac{BC}{FG}=\dfrac{2}{3}

\dfrac{CD}{GH}=\dfrac{22}{34}

\dfrac{CD}{GH}=\dfrac{11}{17}

And,

\dfrac{AD}{EH}=\dfrac{24}{36}

\dfrac{AD}{EH}=\dfrac{2}{3}

Clearly, \dfrac{AB}{EF}=\dfrac{BC}{FG}=\dfrac{AD}{EH}\neq \dfrac{CD}{GH}.

All corresponding sides are not proportional.

Therefore, the figures are neither similar nor congruent. Hence, third option is correct.

7 0
3 years ago
Do you guys know how to do this I am struggling and need help?
Oksanka [162]

Answer:

i think its 234

Step-by-step explanation:

3 0
3 years ago
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