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3 years ago

2k+5=21 answer this problem plz

Mathematics

1 answer:

Anon25 [30]3 years ago

3 0

Answer:

k=8

Step-by-step explanation:

2k+5=21 (-5 on both sides to get rid of +5)

2k=16 (divide by 2 to get rid of that 2 in front of K)

k=8

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What happens to energy that is unused?

Hitman42 [59]

Energy transform to potential energy. as it is stored..I think

3 0

3 years ago

Maxine can swim 240 meters in 4 minutes. What is Maxine's unit rate?

BabaBlast [244]

Answer: A) 18.2 minutes
B) 4.5 minutes.
Step-by-step explanation:
A) Maxine can mow one lawn in 24 minutes, so the rate of work is:
1/24 of the lawn per minute.
Sammie can mow one lanw in 36 minutes, so the rate of work is:
1/36 of the lawn per minute.
The amount of the lawn that each has left to lawn by the minute x is:
Maxine = 1 -(1/24)*x
Sammie = 1 - (1/36)*x
we want to find the value of x such that:
2*(1 - (1/24)*x) = 1 - *(1/36)*x
which means that the amount that Sammie has left is two times the amount that Maxine has left to mow.
2 - (1/12)*x = 1 - (1/36)*x
2 - 1 = (1/12 - 1/36)*x
1/0.055 = x = 18.2
by the minute 18.2
B) Similar to before, here the rates are:
For Maxine, R = 1/6
For Sammie, R = 1/9
The amount they have left to mow by minute x is:
Maxine = 1 - (1/6)*x
Sammie = 1 - (1/9)*x
We want to solve, similar to before.
2( 1 - (1/6)*x) = 1 - (1/9)*x
2 - (1/3)*x = 1 - (1/9)*x
2 - 1 = (1/3 - 1/9)*x
1 = (2/9)*x
1*9/2 = x = 4.5
So here the solution is 4.5 minutes. Hope this helps

5 0

3 years ago

PLEASE HELP ASAP WILL MARK THE BRAINLEST

postnew [5]

Answer:

the correct answer is b

4 0

3 years ago

3x+2y If x=5 and y=2

aev [14]

Plug in the x & y
then multiply
then add
3(5)+2(2)
15+4
19

6 0

3 years ago

Read 2 more answers

You measure 50 textbooks' weights, and find they have a mean weight of 37 ounces. Assume the population standard deviation is 5.

Hoochie [10]

Answer:

90% confidence interval for the true population mean textbook weight is [35.79 ounces , 38.21 ounces].

Step-by-step explanation:

We are given that you measure 50 textbooks' weights, and find they have a mean weight of 37 ounces.

Assume the population standard deviation is 5.2 ounces.

Firstly, the Pivotal quantity for 90% confidence interval for the population mean is given by;

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

where, $\bar X$ = sample mean weight = 37 ounces

$\sigma$ = population standard deviation = 5.2 ounces

n = sample of textbooks = 50

$\mu$ = true population mean textbook weight

Here for constructing 90% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5%

level of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

P( $\bar X -1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X +1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X -1.645 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X +1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $37-1.645 \times {\frac{5.2}{\sqrt{50} } }$ , $37+1.645 \times {\frac{5.2}{\sqrt{50} } }$ ]

= [35.79 , 38.21]

Therefore, 90% confidence interval for the true population mean textbook weight is [35.79 ounces , 38.21 ounces].

7 0

3 years ago