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ehidna [41]
3 years ago
13

Which pair of expressions below are equivalent?

Mathematics
1 answer:
Ede4ka [16]3 years ago
7 0

Answer:

The answer is C

Hope this helped!

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How would you solve a inequality like: 15-13(2x+6)< 15-3(6x-7)
Arturiano [62]
<span>15 - 13(2x + 6) < 15 - 3(6x - 7)
15 - 26x - 78 < 15 - 18x + 21
-26x -63 < -18x + 36
-8x < 99
   x > -99/8
or
   x > - 12.375

</span>
3 0
3 years ago
Read 2 more answers
(-9.1)0= simply this expression
Amanda [17]

Answer:

Well if its -9.1 x 0 that would mean that the answer is 0

Step-by-step explanation:

Use M A T H S P A C E its really helpful

3 0
3 years ago
Mary borrowed $700 from a bank for 3 years and was charged simple interest. The total interest that she paid on the loan $42. As
Mashcka [7]

Answer:

2 %

Step-by-step explanation:

Money borrowed, Principal, P = $ 700

time of investment, T = 3 years

Total interest paid, Simple interest, S.I = $ 42

Let the rate of interest is R.

Use the formula for simple interest.

S.I. =\frac{P\times R\times T}{100}

By substituting the values

42 =\frac{700\times R\times 3}{100}

4200 = 2100 R

R = 2 %

Thus, the rate of interest is 2%.

3 0
4 years ago
What is the area of a rectangle with a length of 1/2 yard and a width of 11/6
Nataly [62]
56.................................................
7 0
4 years ago
the patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.9 days and a standard de
NNADVOKAT [17]
The 85th percentile is the cutoff time t such that

\mathbb P(X

In other words, the 85th percentile refers to the time needed to belong to the top 15% of the distribution; more generally, the n percentile is the top (100-n)\% of the distribution.

Anyway, to find this value of t, transform X to a random variable Z with the standard normal distribution using

Z=\dfrac{X-\mu}\sigma

where \mu is the mean of X and \sigma is the standard deviation of X.

\mathbb P(X

Here t^* is used to denote the z-score corresponding to the cutoff time t. Referring to a z-score table, you find that this occurs for t^*\approx1.036. So,

\dfrac{t-5.9}{2.1}=t^*\implies t=5.9+2.1t^*\approx8.076
7 0
3 years ago
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