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mafiozo [28]
3 years ago
12

NEED THE ANSWER ASAP PLEASE

Mathematics
1 answer:
Juliette [100K]3 years ago
4 0
The answer is y=1x+7
1 4
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Bro help me idk this one q-q <br><br>n × 5 + 4 × 2 - 4 × 3 = 42​
lina2011 [118]

Answer:

n = 9.2

Step-by-step explanation:

Isolate the variable by dividing each side by the numbers that do not contain the variable

7 0
3 years ago
Please please answer this correctly. Please take a pic of the same screenshot of put the right plots correctly
sergeinik [125]

Answer:

Picture

Step-by-step explanation:

I graphed them

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3 years ago
Hi i am having trouble dividing decimals, im not really sure how to devide at all i keep forgeting.
kolbaska11 [484]
If You know how to do long division then... You move the decimal that’s outside the house to where it’s not there no more then divide . Or turn the decimals into Fractions and do the reciprocal.
5 0
3 years ago
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
3 years ago
Raj recorded the scores of six of his classmates in the table.
WINSTONCH [101]
98-70 is 28 meaning that the range is 28
4 0
3 years ago
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