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balandron [24]
3 years ago
10

Find the solution set of this inequality. select the correct graph. |-4x+14|>6

Mathematics
1 answer:
lions [1.4K]3 years ago
5 0
Split the inequality into two:
(-4x+14)>6 & -(-4x+14)>6

solve:
(-4x+14)>6
-4x>-8
x<2
(symbol changes when ÷/× by negative for inequalities.)

-(-4x+14)>6
-4x+14<-6
-4x<-20
x>5

the answers are x<2 and x>5, so you would write your answer like this: (-∞,2) ∪ (5,∞)
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Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.8 parts
Alexxandr [17]

Answer:

The value of the test statistic is t = 2.67

Step-by-step explanation:

The null hypothesis is:

H_{0} = 7.8

The alternate hypotesis is:

H_{1} \neq 7.8

Our test statistic is:

t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

In this problem, we have that:

X = 8.2, \mu = 7.8, \sigma = 0.6, n = 16

Then

t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

t = \frac{8.2 - 7.8}{\frac{0.6}{\sqrt{16}}}

t = 2.67

The value of the test statistic is t = 2.67

5 0
3 years ago
The points ​(​15, 18​) and ​(​35,42) form a proportional relationship. Find the slope of the line through the points. Then use t
Ann [662]

Answer:

6/5

Step-by-step explanation:

(42-18) / (35 - 15)  = 24 / 20  = 6 / 5

8 0
3 years ago
Spam e-mail containing a virus is sent to 1000 e-mail addresses. After 1 second, a recipient machine broadcasts 10 new spam e-ma
pickupchik [31]

Answer:

Answer explained below

Step-by-step explanation:

Spam e-mail containing a virus is sent to 1000 e-mail addresses. After 1 second, a recipient machine broadcasts 10 new spam e-mails containing the same virus, after which the virus disables itself on that machine. (1) Write a recursive definition (i.e. recurrence relation) to show how many spam emails will be sent out after n seconds. (2) Solve the recurrence relation. (3) How many e-mails are sent at the end of 20 seconds

1.START T=0.....

1000 EMAILS SENT AND RECEIVED BY 1000 M/CS.

T=1.....

EACH OF THE M/C SENDS 10 NEW MAILS ....

.................................

LET M[N] BE THE NUMBER OF MAILS SENT OUT AFTER N SECONDS.

SO , EACH OF THESE M/CS WILL SEND 10 MAILS IN NEXT 1 SECOND.

HENCE NUMBER OF MAILS SENT IN N+1 SECONDS=M[N+1]=

M[N+1]=10*M[N].......................1

THIS IS THE RECURRENCE RELATION.....

2.SOLUTION .....

M[N+1]=10M[N]=10*10M[N-1]=10*10*10M[N-2]=........

M(N+1)=[10^1][M(N)]=[10^2][M(N-1)]=[10^3][M(N-2)]=..........=[10^N][M(1)]=[10^(N+1)][M(0)]

M[N+1]=[10^(N+1)][1000]=[10^(N+1)][10^3]=[10^(N+4)]......................................2

THIS IS THE NUMBER OF MAILS SENT AFTER N+1 SECONDS .....OR ....

M[N]=[10^(N+3)].............................................3

..................IS THE SOLUTION FOR NUMBER OF MAILS SENT AFTER N SECONDS.....

3.AFTER N=20 SECONDS , THE ANSWER IS ....

M[20]=10^(20+3)=10^23

8 0
3 years ago
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77julia77 [94]
No he is not, 10*0.13=1.3, so 5*0.13=0.65
6 0
3 years ago
Read 2 more answers
Which algebraic expression represents this phrase? the product of 86 and the depth of the river
Marizza181 [45]
If "d" is used to represent the depth of the river, that product will be
  86d
6 0
3 years ago
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