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oksano4ka [1.4K]
3 years ago
9

What is d in 6d− 2/11 =2d− 2/13? ​

Mathematics
1 answer:
Julli [10]3 years ago
6 0

Answer:

-1/4 or -0.25

Step-by-step explanation:

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The table and the Circle Chart shown display the percentage of dogs in seven different groups of dog breeds in a dog competition
Varvara68 [4.7K]

Answer:

a

Step-by-step explanation:

4 0
2 years ago
Please i need help rn​
Aleks04 [339]

Answer:

here u go

Step-by-step explanation:

16 /5      =3 and 1 over 53  1 /5=3.2

8 0
3 years ago
What is the profitability index of an investment with cash flows in years 0 thru 4 of -340, 120, 130, 153, and 166, respectively
ki77a [65]

The profitability index of an investment with cash flows in years 0 thru 4 of -340, 120, 130, 153, and 166, respectively, and a discount rate of 16 percent is: 15%.

<h3>Profitability index</h3>

First step is to find the Net present value (NPV)  of the given cash flow using discount rate PVF 16% and PV of cash flow which in turn will give us net present value of 49.7.

Second step is to calculate the profitability index

Profitability index  = 49.7/340

Profitability index  = .15×100

Profitability index=15%

Therefore the profitability index of an investment with cash flows in years 0 thru 4 of -340, 120, 130, 153, and 166, respectively, and a discount rate of 16 percent is: 15%.

Learn more about Profitability index here:brainly.com/question/3805108

#SPJ4

3 0
1 year 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
If 5-3(2a+1)=-4a+10, what is the value of a+2​
AlekseyPX

Answer:

-2

<em>BRAINLIEST, PLEASE!</em>

Step-by-step explanation:

5 - 3(2a + 1) = -4a + 10

5 - 6a - 3 = -4a + 10

2 - 6a = -4a + 10

-2a = 8

a = -4

-4 + 2 = -2

5 0
3 years ago
Read 2 more answers
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