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andre [41]
2 years ago
15

Which of the following equations have exactly one solution? Choose all answers that apply:

Mathematics
1 answer:
MAXImum [283]2 years ago
8 0

Answer:

C

Step-by-step explanation:

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A student says that the​ x-intercept of a line is ​(0,-10,​). Is the student​ correct? Explain.
nataly862011 [7]

Answer:

Yes, the student is correct

Step-by-step explanation:

Given

(0,-10)

Required

Does it represent the x intercept

At the x intercept, the value of x is 0.

So, the function (x,y) becomes (0,y)

(0,-10) can be compared to (0,y)

Hence, the student is correct

5 0
2 years ago
You inherit $5000 from your long lost Uncle Harold. The bad news is that the money must sit in a bank account for the next ten y
erastovalidia [21]

Answer:

5360

Step-by-step explanation:

5000*1.072=5360

6 0
1 year ago
PLS HELPPP MEEE ILL GIVE BRAINLIEST
Arlecino [84]

Answer:

y = 2

Step-by-step explanation:

not sure but should be it

4 0
3 years ago
Read 2 more answers
Express -5 1/4 as a decimal
rewona [7]
I believe it would be -5.25
5 0
3 years ago
Read 2 more answers
The amount of time a passenger waits at an airport check-in counter is random variable with mean 10 minutes and standard deviati
Stolb23 [73]

Answer:

(a) less than 10 minutes

= 0.5

(b) between 5 and 10 minutes

= 0.5

Step-by-step explanation:

We solve the above question using z score formula. We given a random number of samples, z score formula :

z-score is z = (x-μ)/ Standard error where

x is the raw score

μ is the population mean

Standard error : σ/√n

σ is the population standard deviation

n = number of samples

(a) less than 10 minutes

x = 10 μ = 10, σ = 2 n = 50

z = 10 - 10/2/√50

z = 0 / 0.2828427125

z = 0

Using the z table to find the probability

P(z ≤ 0) = P(z < 0) = P(x = 10)

= 0.5

Therefore, the probability that the average waiting time waiting in line for this sample is less than 10 minutes = 0.5

(b) between 5 and 10 minutes

i) For 5 minutes

x = 5 μ = 10, σ = 2 n = 50

z = 5 - 10/2/√50

z = -5 / 0.2828427125

= -17.67767

P-value from Z-Table:

P(x<5) = 0

Using the z table to find the probability

P(z ≤ 0) = P(z = -17.67767) = P(x = 5)

= 0

ii) For 10 minutes

x = 10 μ = 10, σ = 2 n = 50

z = 10 - 10/2/√50

z = 0 / 0.2828427125

z = 0

Using the z table to find the probability

P(z ≤ 0) = P(z < 0) = P(x = 10)

= 0.5

Hence, the probability that the average waiting time waiting in line for this sample is between 5 and 10 minutes is

P(x = 10) - P(x = 5)

= 0.5 - 0

= 0.5

3 0
3 years ago
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