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GarryVolchara [31]
3 years ago
9

What is the product of 2.5 × 10−15 and 3.9 × 1026?

Mathematics
1 answer:
miskamm [114]3 years ago
6 0

Answer:

40014

Step-by-step explanation:

2.5 × 10=25                                                    3.9 × 1026=

25-15=                                                          4001.4  

10


4001.4 × 10= 40014


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sergij07 [2.7K]

Answer:

Yes you are right on both.

Step-by-step explanation:

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3 years ago
. If mZJ= (7x + 13)", mZK = (83 - 2x), and
OLga [1]

Answer:

  • m∠J = 76°
  • m∠K = 65°

Step-by-step explanation:

The sum of angle measures is ...

  (7x +13)° +(83 -2x)° = 141°

  5x +96 = 141 . . . . . collect terms, divide by °

  5x = 45 . . . . . . . . . subtract 96

  x = 9 . . . . . . . . . . . divide by 5

  7x +13 = 7·9 +13 = 76 . . . . . find m∠J

  m∠J = 76°

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3 years ago
What does x equal in -18= 15 minus 3(6x + 5)
aalyn [17]

Answer:

x = -1

Step-by-step explanation:

-18=15-3(6x+5)              Distributive Property

-18=15-18x-15                Combining like terms.

-18=18x                          Divide by the coefficient.

-1=x

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3 years ago
Please help please ill give brainliest best answer
Molodets [167]

The first step is to find the area of the figure amd then multiply it by the weight of per metre...

<h3>Step 1</h3>

  • Area of figure⤵️

\sf \: area = length \times width

  • Length = 60 cm = 0.6 metre
  • Width = 200 cm = 2 metre

\bf \: area = 0.6 \times 2

\bf \: area = 1.2 \:  {m}^{2}

<h3>Now,</h3>

  • Weight of 1 metre² = 0.9 kg

<h3>So, The weight of 1.2 m² is; </h3>

\tt \: 1.2 \:{m}^{2}  = 1.2 \: kg

The framework is made up of 5 rods, So the weight will be

  • 1.2 × 5 = 6kg

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3 years ago
On average 20% of the gadgets produced by a factory are mildly defective. I buy a box of 100 gadgets. Assume this is a random sa
Softa [21]

Answer:

a) P(A)=P(x

b) P(A)\approx 0.085

Step-by-step explanation:

The proportion of defective gadgets for this factory is p=0.2.

If a sample of n=100 gadgets is taken from the production, we can model the amount of defectives gadgets in this sample as a binomial random variable.

The probability of having k defectives in the sample can be written as:

P(x=k)=\binom{100}{k}p^k(1-p)^{100-k}

Then, we can write the expression to calculate the probabilities of A: "less than 15 gadgets are mildly defective":

P(A)=P(x

If we approximate this binomial distribtution to a normal distribution, we can calculate the new parameters as:

\mu=np=100*0.2=20\\\\\sigma=\sqrt{np(1-p)}=\sqrt{100*0.2*0.8}=\sqrt{16}=4

The continuity factor is applied for the change from a discrete distribution (binomial) to a continous distribution (normal). Then we have:

P(A)=P(X

Now, we can calculate the probability P(A):

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