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snow_tiger [21]
3 years ago
15

Does anyone know how to solve one of these ?

Mathematics
1 answer:
BigorU [14]3 years ago
6 0
Sorry, But no. I hope you figure it out :)

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F (n)=-3n^2-4n<br> Find f(4)
defon

Answer:

144

Step-by-step explanation:

f(4)=3*4^2

f(4)=12^2

12*12=144

7 0
3 years ago
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Help urgently please!
Nataliya [291]
1.
Slope-300.
Y int: 40,000
Y=300x+40000
2.
S: -5
Y: 250
Y=-5x+250
3.
S: 2
Y: 5
Y=2x+5
4.
S: 150
Y: 350
Y=150x+350
The last question is cut off.
5 0
3 years ago
Two Trains left a station at the same time. One traveled north at a certain speed and the other traveled south at twice that spe
sashaice [31]
Let s = northbound train
Then
2s = southbound train
:
Distance = time * speed
4s + 4(2s) = 600
:
4s + 8s = 600
:
12s = 600
:
s = 600/12
:
s = 50 mph is the northbound train
Then
2(50) = 100 mph is the southbound train
:
:
Check:
4(50) + 4(100) = 600
3 0
3 years ago
Read 2 more answers
The amount of time that people spend at Grover Hot Springs is normally distributed with a mean of 73 minutes and a standard devi
Vesnalui [34]

Answer:

(a) X\sim N(\mu = 73, \sigma = 16)

(b) 0.7910

(c) 0.0401

(d) 0.6464

Step-by-step explanation:

Let <em>X</em> = amount of time that people spend at Grover Hot Springs.

The random variable <em>X</em> is normally distributed with a mean of 73 minutes and a standard deviation of 16 minutes.

(a)

The distribution of the random variable <em>X</em> is:

X\sim N(\mu = 73, \sigma = 16)

(b)

Compute the probability that a randomly selected person at the hot springs stays longer than 60 minutes as follows:

P(X>60)=P(\frac{X-\mu}{\sigma}>\frac{60-73}{16})\\=P(Z>-0.8125)\\=P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that a randomly selected person at the hot springs stays longer than an hour is 0.7910.

(c)

Compute the probability that a randomly selected person at the hot springs stays less than 45 minutes as follows:

P(X

*Use a <em>z</em>-table for the probability.

Thus, the probability that a randomly selected person at the hot springs stays less than 45 minutes is 0.0401.

(d)

Compute the probability that a randomly person spends between 60 and 90 minutes at the hot springs as follows:

P(60

*Use a <em>z</em>-table for the probability.

Thus, the probability that a randomly person spends between 60 and 90 minutes at the hot springs is 0.6464

6 0
3 years ago
WILL MARK BRAINLYIST IF CORRECT!
Nady [450]
Your answer will be C
6 0
3 years ago
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