All categories

2 years ago

Can somebody please help me with this question??10=10(k-9)

Mathematics

2 answers:

dmitriy555 [2]2 years ago

5 0

answer:

k = 10

step-by-step explanation:

objective: get k by itself on one side of the equal sign

1. distribute the 10 to (k - 9) by multiplying these values by the outside value

10 × k = 10k

10 × -9 = -90

new equation: 10 = 10k - 90

2. get rid of 90 by doing the opposite (so we add 90) and what we do on one side we do to the other

(10k - 90) + 90 = 10k (90 cancels out)

10 + 90 = 100

new equation: 100 = 10k

3. get k by itself by dividing by 10 - since 10 is being multiplied by we need to get rid of it by doing its opposite (division) and doing this on both sides

10k ÷ 10 = k (the 10 cancels out)

100 ÷ 10 = 10

new equation: 10 = k

is k by itself? yes so k is equal to 10

SashulF [63]2 years ago

4 0

K = 10

I think. I’m not positive sure though. Hope this helped!

You might be interested in

What is the greatest common factor of 24s^3 , 12s^4 , and 18s

denpristay [2]

6s. 2 and 3 divide into all these numbers but six is the greatest number that all
Divide into and they all have one s at least.

6 0

2 years ago

Read 2 more answers

Heplllllll on 2 3 4 5

sveticcg [70]

Change them into improper fractions
and then find a common denominator, then minus, add, whatever for each one

7 0

2 years ago

For f(x)=4x+1 and g(x)=x^2-5,find(f•g)(4)

RoseWind [281]

Answer:

(f•g)(4) = 45

Step-by-step explanation:

f(x)=4x+1

g(x)=x^2-5

(f•g)(x) = 4(x^2 -5)+1

(f•g)(4) = 4(4^2 -5)+1

(f•g)(4) = 4(16-5)+1

(f•g)(4) = 4(11)+1

(f•g)(4) = 44 + 1

(f•g)(4) = 45

8 0

2 years ago

Read 2 more answers

Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis

lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = the interval of time between the eruption

So, X ~ Normal( $\mu=75, \sigma^{2} =20$ )

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

P(X > 84 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{84-75}{20}$ ) = P(Z > 0.45) = 1 - P(Z $\leq$ 0.45)

= 1 - 0.6736 = 0.3264

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{13} } }$ ) = P(Z > 1.62) = 1 - P(Z $\leq$ 1.62)

= 1 - 0.9474 = 0.0526

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{20} } }$ ) = P(Z > 2.01) = 1 - P(Z $\leq$ 2.01)

= 1 - 0.9778 = 0.0222

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

8 0

2 years ago

Which linear function represents a slope of 1?

algol13

Yes yes woeoshdenenejdjdjdjejejrjrrjekekekekekeek 30

8 0

2 years ago

Read 2 more answers