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lutik1710 [3]
2 years ago
6

*** Please explain your answer***

Mathematics
1 answer:
Helen [10]2 years ago
7 0
$49.30 (72.50*.20)=58 

58*.15 = 49.30
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natka813 [3]
\frac{z}{3}+ \frac{12}{3}= \frac{5}{6} ⇒ \frac{z+12}{3}= \frac{5}{6}

z+12= \frac{5\cdot3}{6}=\boxed{2.5}

z=2.5-12

\boxed{\boxed{\boxed{z=-9.5}}}
4 0
3 years ago
Read 2 more answers
If g(x-1) = (2-g(x))/8 and g(1) = 3, find the value of g(2).
aniked [119]

Answer:

-22

Step-by-step explanation:

g(2-1)=g(1)=(2-g(2))/8=3

2-g(2)=24

-g(2)=22

g(2)=-22

5 0
3 years ago
Please answer the question please. The question is down below.
nekit [7.7K]

Answer:

Please add a picture!

Step-by-step explanation:

7 0
3 years ago
The time it takes a student to walk from the dorm to the chemistry lab follows roughly a Normaldistribution with a mean of 20 mi
creativ13 [48]

Answer:

95.15%

Step-by-step explanation:

We have that the mean (m) is equal to 20, the standard deviation (sd) 3

They ask us for P (x <25)

For this, the first thing is to calculate z, which is given by the following equation:

z = (x - m) / sd

We have all these values, replacing we have:

z = (25 - 20) / (3)

z = 1.66

With the normal distribution table (attached), we have that at that value, the probability is:

P (z <1.66) = 0.9515

Which means that the probability that it arrives before 25 minutes is 95.15%

6 0
3 years ago
Determine whether the following statements are True or False.
Stella [2.4K]

Answer:

1) False

2) False

3) True

4) False

Step-by-step explanation:

1) Flase, {v1,v2,v3, ..., vp} is a base for H when they span H and also they are linearly independent.

2) False. A single nonzero vector is linearly independent , not dependent. There is not null linear combination that gives 0 as a result involving that vector.

3) True, if the columns werent linearly independent, we could triangulate the matrix and obtain 0, so the matrix wouldnt be invertible. This means that the columns should be linearly independent for the matrix to be invertible and as a consecuence, they will spam a subspace of R^n of dimension n, which means that they will spam all R^n and therefore, they form a basis of R^n.

4) False. A basis is a spanning set that is as small as possible. Larger spanning sets will have extra elements apart from those who can form a base toguether. Those elements will make the set linearly dependent.

6 0
2 years ago
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