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Bogdan [553]
3 years ago
10

WHICH INQUALITY IS EQUAL TO -1 > M+4

Mathematics
1 answer:
PIT_PIT [208]3 years ago
5 0

Answer:

m<−5

Step-by-step explanation:

Step 1: Flip the equation.

m+4<−1

Step 2: Subtract 4 from both sides.

m+4−4<−1−4

m<−5

Hope this helped <3

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The number 160 is increased by 50%. The result is then decreased by 25%. What is the final number?
harkovskaia [24]

Answer:

60

Step-by-step explanation:

50%=.50

160×.50=80

160+80=240

25%=.25

240×.25=60

60 is the final number

8 0
2 years ago
What is the output when the input is 0
Cloud [144]

Answer:

When the inputs are 1 and 0, the output is zero.

7 0
3 years ago
Please answer this to the best of your ability
sweet [91]

f(-2)=1.5

f(0)=0.5

f(3)=-1

:)

7 0
3 years ago
The manager went over the sales of mobile phones at the store and found that the mean sale was 45, with a standard deviation of
d1i1m1o1n [39]

The z-score of the sale of mobile phones on that day is 1.75

Step-by-step explanation:

The formula of z-score is z = (x - μ)/σ, where:

  • x is the score
  • μ is the mean
  • σ is the standard deviation

∵ The mean sale was 45

∴ μ = 45

∵ The standard deviation was 4

∴ σ = 4

∵ 52 mobile phones were sold on a particular day

∴ x = 52

To find z-score of the sale of mobile phones on that day substitute the values of x, μ, and σ in the formula of z-score

∵ z=\frac{52-45}{4}

∴ z=\frac{7}{4}

∴ z = 1.75

The z-score of the sale of mobile phones on that day is 1.75

Learn more:

You can learn more about z-score in brainly.com/question/7207785

#LearnwithBrainly

3 0
3 years ago
A certain substance doubles its volume every minute at 9 AM a small amount is placed in a container ,at 10 AM the container was
Y_Kistochka [10]

Answer:

the container is 1/4 full at 9:58 AM

Step-by-step explanation:

since the volume doubles every minute , the formula for calculating the volume V at any time t is

V(t)=V₀*2^-t , where t is in minutes back from 10 AM and V₀= container volume

thus for t=1 min (9:59 AM) the volume is V₁=V₀/2 (half of the initial one) , for t=2 (9:58 AM) is V₂=V₁/2=V₀/4  ...

therefore when the container is 1/4 full the volume is V=V₀/4 , thus replacing in the equation we obtain

V=V₀*2^-t

V₀/4 = V₀*2^-t

1/4 = 2^-t

appling logarithms

ln (1/4) = -t* ln 2

t = - ln (1/4)/ln 2 = ln 4 /ln 2 = 2*ln 2 / ln 2 = 2

thus t=2 min before 10 AM → 9:58 AM

therefore the container is 1/4 full at 9:58 AM

8 0
2 years ago
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