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sdas [7]
3 years ago
8

Answer please I need real bad help and I live alone

Mathematics
2 answers:
tatyana61 [14]3 years ago
8 0

Answer:

C is the answer

Step-by-step explanation:

VARVARA [1.3K]3 years ago
6 0

Answer:

The answer is C

Step-by-step explanation:

First you look to the X axis and then the Y, so you found the -3 at the X axis and you look to the Y axis. Then you try and find the 4, so the answer is C. Remember to always look at the X axis first.

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GuDViN [60]

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3. QU = QU 3. Reflexive

4. QA = UD 4. Diagonals of an isosceles trapezoid are equal.

5. ΔUAQ = ΔQDU 5. SSS

6. ∠UAQ = ∠QDU 6. CPCTE

3 0
3 years ago
Help please!!!! 10 points
Artemon [7]

Step-by-step explanation:

This is the answer.... Working shown

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3 years ago
How to do property eqautionseqautions
Tanya [424]

Answer:


Step-by-step explanation:


4 0
3 years ago
In ΔUVW, w = 9 cm, v = 2.2 cm and ∠V=136°. Find all possible values of ∠W, to the nearest 10th of a degree.
Alja [10]

Complete Question

In ΔUVW, w = 9 cm, v = 22 cm and ∠V=136°. Find all possible values of ∠W, to the nearest 10th of a degree.

Answer:

16.5°

Step-by-step explanation:

In ΔUVW, w = 9 cm, v = 22 cm and ∠V=136°. Find all possible values of ∠W, to the nearest 10th of a degree.

We solve using Sine rule formula

a/sin A = b/sin B

We are solving for angle W

∠V=136°

Hence:

22 /sin 136 = 9 /sin W

Cross Multiply

22 × sin W = sin 136 × 9

sin W = sin 136 × 9/22

W = arc sin [sin 136 × 9/2.2]

W = 16.50975°

W = 16.5°

3 0
3 years ago
A manager at a local company asked his employees how many times they had given blood in the last year. The results of the survey
Lubov Fominskaja [6]

Answer:

Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779

And the deviation would be:

Sd(X) =\sqrt{2.3779}= 1.542 \approx 1.54

Step-by-step explanation:

For this case we have the following distribution given:

X        0         1       2       3       4         5        6

P(X)  0.3   0.25   0.2   0.12   0.07   0.04   0.02

For this case we need to find first the expected value given by:

E(X) = \sum_{i=1}^n X_i P(X_I)

And replacing we got:

E(X)= 0*0.3 +1*0.25 +2*0.2 +3*0.12 +4*0.07+ 5*0.04 +6*0.02=1.61

Now we can find the second moment given by:

E(X^2) =\sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2)= 0^2*0.3 +1^2*0.25 +2^2*0.2 +3^2*0.12 +4^2*0.07+ 5^2*0.04 +6^2*0.02=4.97

And the variance would be given by:

Var(X) = E(X^2) -[E(X)]^2 = 4.97 -(1.61)^2 =2.3779

And the deviation would be:

Sd(X) =\sqrt{2.3779}= 1.542 \approx 1.54

 

8 0
3 years ago
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