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iren2701 [21]
4 years ago
10

1 ) Alyssa has 2 blue marbles Jason has 6 blue marbles.

Mathematics
1 answer:
s2008m [1.1K]4 years ago
8 0

Answer:

The problem only includes blue marbles.

Step-by-step explanation:

Lets visualize this.

(the O represents marbles)

Alyssa has 2 marbles -->  O O

Jason has 6 marbles -->   O O O O O O

Let's count them. How many marbles do we have all together?

O  O  O  O  O  O  O  O

1    2   3  4   5   6   7   8

So that's 8 marbles! You have 8 total marbles.

I hope this helps!

<h2><u>PLEASE MARK BRAINLIEST!</u></h2>
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2 years ago
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In the figure CD is the perpendicular bisector of AB . If the length of AC is 2x and the length of BC is 3x - 5 . The value of x
Dennis_Churaev [7]

Answer

Find out the value of x .

To proof

SAS congurence property

In this property two sides and one angle of the two triangles are equal.

in the Δ ADC and ΔBDC

(1) CD = CD (common side of both the triangle)

(2) ∠CDA = ∠ CDB = 90 °

( ∠CDA +∠ CDB = 180 ° (Linear pair)

as given in the diagram

∠CDA  = 90°

∠ CDB = 180 ° - 90°

∠ CDB = 90°)

(3) AD = DB (as shown in the diagram)

Δ ADC ≅ ΔBDC

by using the SAS congurence property .

AC = BC

(Corresponding sides of the congurent triangle)

As given

the length of AC is 2x and the length of BC is 3x - 5 .

2x = 3x - 5

3x -2x =5

x = 5

The value of x is 5 .

Hence proved


7 0
3 years ago
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Consider the sequence: 11, 21, 31, …, … , … What is the nth term of this sequence ?
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Criterion: Identify IV, DV, and hypotheses and evaluate the null hypothesis for an independent samples t test. Data: Use the inf
Papessa [141]

The table is missing in the question. The table is provided here :

Group 1        Group 2

 34.86            64.14      mean

 21.99            20.46      standard deviation

  7                    7                n

Solution :

a). The IV or independent variable = Group 1

    The DV or the dependent variable = Group 2

b).

  $H_0: \mu_1 = \mu_2$

  $H_a:\mu_1 < \mu_2$

  Therefore,   $t = \frac{\bar x_1 - \bar x_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{S_2^2}{n_2}}}$

$t = \frac{34.86 - 64.14}{\sqrt{\frac{21.99^2}{7}+\frac{20.46^2}{7}}}$

t = -2.579143

Now,   $df = min(n_1 - 1, n_2 - 1)$

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               = 6

Therefore the value of p :

  $=T.DIST(-2.579143,6,TRUE)$

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The p value is 0.0209

$p< 0.05$

So we reject the null hypothesis and conclude that $\mu_1 < \mu_2$

7 0
3 years ago
Mrs. Barrera learned that 21 of
otez555 [7]

Answer:

just do 21/25

Step-by-step explanation:

8 0
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