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Elena L [17]
3 years ago
7

Can someone answer these 2 questions please? Thanks

Mathematics
1 answer:
natali 33 [55]3 years ago
3 0
The first one is -12 the second is 10
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Can someone help me please
fredd [130]
The answer to that is C.
6 0
3 years ago
Solve the inequality 171>-6x and graph the solution what does the graph look like
krok68 [10]

To solve the inequality, you need to isolate/get x by itself:

171 > -6x    Divide -6 on both sides [dividing/multiplying a negative number on

-28.5 < x      [dividing/multiplying a negative number in an inequality causes the sign (<, >, ≤, ≥) to flip]

-28.5 < x    [x is a number greater than -28.5]

So your graph should have an open circle at -28.5 (the first small line next to -28), and the arrow pointing to the right since x is greater than -28.5 (increasing)      The 1st option is your answer

[use the o---> and put it at -28.5]

7 0
4 years ago
When adding or subtracting to decimals what is the first thing you must do
Svetach [21]
Line up the decimals 
3 0
3 years ago
Geometry help will give brainliest
blsea [12.9K]
In a parallogram the two angles on the same side ( Angle T and Angle C) equal 180 degrees

So we have 8x +29 + 2x +11 = 180

combine the like terms:
10x + 40 = 180

Subtract 40 from each side:
10x = 140

Divide each side by 10:
X = 140 /10
X = 14

Now we have X, replace X into the equation for angle C

2(14) +11 = 28 + 11 = 39 degrees

6 0
3 years ago
Read 2 more answers
Resistors are labeled 100 Ω. In fact, the actual resistances are uniformly distributed on the interval (95, 103). Find the mean
Zinaida [17]

Answer:

E[R] = 99 Ω

\sigma_R = 2.3094 Ω

P(98<R<102) = 0.5696

Step-by-step explanation:

The mean resistance is the average of edge values of interval.

Hence,

The mean resistance, E[R] = \frac{a+b}{2}  = \frac{95+103}{2} = \frac{198}{2} = 99 Ω

To find the standard deviation of resistance, we need to find variance first.

V(R) = \frac{(b-a)^2}{12} =\frac{(103-95)^2}{12} = 5.333

Hence,

The standard deviation of resistance, \sigma_R = \sqrt{V(R)} = \sqrt5.333 = 2.3094 Ω

To calculate the probability that resistance is between 98 Ω and 102 Ω, we need to find Normal Distributions.

z_1 = \frac{102-99}{2.3094} = 1.299

z_2 = \frac{98-99}{2.3094} = -0.433

From the Z-table, P(98<R<102) = 0.9032 - 0.3336 = 0.5696

5 0
4 years ago
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