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valentinak56 [21]
3 years ago
14

6th grade math plz help me

Mathematics
1 answer:
vlabodo [156]3 years ago
5 0

Answer:

The mode, which is 7, is not a typical distance for the data. The median, which is 3.5, is not a typical distance for the data. The mean, which is 4, is a typical distance for the data. So, Grace’s reasoning is not correct.

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If cosA=3√2/5,then show that cos2A=11/25
Kobotan [32]

Answer:

Step-by-step explanation:

Cos 2A = 2Cos² A - 1

             = 2*(\frac{3\sqrt{2}}{5})^{2}-1\\\\=2*(\frac{3^{2}*(\sqrt{2})^{2}}{5^{2}})-1\\\\=2*\frac{9*2}{25} - 1\\\\=\frac{36}{25}-1\\\\=\frac{36}{25}-\frac{25}{25}\\\\=\frac{11}{25}

4 0
3 years ago
Please help me please
icang [17]

Answer:

change/original price X 100

22/88 X 100 = 25%

4 0
3 years ago
Consider the population of four juvenile condors. Their weights in pounds are : 4, 5, 7, 12 (a) Let x be the weight of a juvenil
creativ13 [48]

Answer:

a) 4, 5, 7, 12

b) 7

c) 4.5, 6.5, 8, 6, 8.5, 9.5

d) 7.167              

Step-by-step explanation:

We are given the following in the question:

4, 5, 7, 12

a) unique values for x

4, 5, 7, 12

b) mean of the population

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

\mu =\displaystyle\frac{28}{4} = 7

c) sampling distribution for samples of size 2

Sample size, n = 2

Possible samples of size 2 are (4,5),(4,7),(4,12),(5,7),(5,12),(7,12)

Sample means are:

\bar{x_1} = \dfrac{4+5}{2} = 4.5\\\\\bar{x_2} = \dfrac{4+7}{2} = 6.5\\\\\bar{x_3} = \dfrac{4+12}{2} = 8\\\\\bar{x_4} = \dfrac{5+7}{2} = 6\\\\\bar{x_5} = \dfrac{5+12}{2} = 8.5\\\\\bar{x_6} = \dfrac{7+12}{2} = 9.5

Thus, the list is 4.5, 6.5, 8, 6, 8.5, 9.5

d) mean of the sampling distribution

\bar{x} = \dfrac{4.5 + 6.5 + 8+ 6 + 8.5+ 9.5}{6} = \dfrac{43}{6} = 7.167

6 0
3 years ago
Could someone please help me with this problem?<br>I keep getting the wrong answer
ahrayia [7]
When given an angle, the side opposite the angle, and another side, you have to determine how many triangles are possible or if it is not possible.

Begin with when a triangle cannot exist: 
1.) measure of non-inclusive angle is less than 90° and side opposite the angle < (other side)×sin(angle measure)
2.) measure of non-inclusive angle is ≥ 90° and side opposite the angle < other side

Next determine if two triangles exist:
Only if the measure of angle is less than 90° and 
(other side)×sin(non-inclusive angle measure) < opposite side < other side

Otherwise 1 triangle exist...
HERE IS WHAT YOUR PROBLEM HAS:
Non-inclusive angle measure = 22° which is < 90°
Opposite side = 13
Other side = 18.1
(other side)×sin(non-inclusive angle measure) =(18.1)×sin(22°) ≈ 6.78

So how many triangles?
6.78 < 13 < 18.1 so 2 triangles exist

Now let's find them... find angle B with law of sines
\frac{sin22^o}{13} = \frac{sinB}{18.1}

Put the following in your calculator: sin^{-1}( \frac{18.1sin(22^o)}{13})
This gives you the first angle B value and if you subtract it from 180 you get the other angle B value

To find the angle C for the first triangle 180 - (sum of angle A and first angle B)
Then use the law of sines to find side c for first triangle.

To find the angle C for the 2nd triangle 180 - (sum of angle A and 2nd angle B)
Then use the law of sines to find side c for 2nd triangle.

Sorry, I didn't do the calculations because my calculator is dead.

 
4 0
3 years ago
Maurice spent 1/2 of his money on lunch. He had 2.50 left.<br> How much money did he start with?
OverLord2011 [107]

Answer:

5.0

Step-by-step explanation:

1÷2=2.5

1=2.50×2

=5.0

6 0
3 years ago
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