Ask question

Ask question

All categories

3 years ago

9

A function g models a relationship in which the dependent variable is multiplied by 4 for every 2 units the independent variable

increases. The value of the function at 0 is 5. Write an equation that represents the function.

1 answer:

12345 [234]3 years ago

6 0

2x+5=y this should be correct.

You might be interested in

Jenny and Hanan are collecting clothes for a clothing drive. Hanan collected 1/2 as many bags of clothes as Jenny did. If Jenny

Montano1993 [528]

Answer:

Hanan collect $\frac{3}{8}$ of a bag of clothes

Step-by-step explanation:

Let

x ------> amount of clothing bag that Jenny collected

y ------> amount of clothing bag that Hanan collected

we know that

$y=\frac{1}{2}x$ -----> equation A

$x=\frac{3}{4}$ -----> equation B

Substitute equation B in equation A

$y=\frac{1}{2}(\frac{3}{4})$

$y=\frac{3}{8}$

therefore

Hanan collect $\frac{3}{8}$ of a bag of clothes

5 0

3 years ago

Wewaii [24]

Answer:

For covering $1$ unit area of the entire playground, the amount of sand required is equal to volume of $3$ buckets of sand.

Step-by-step explanation:

Given -

$\frac{1}{3}$ volume of sand in bucket is able to cover $\frac{1}{9}$ area of the entire playground

Thus,

For covering $\frac{1}{9}$ unit area of the entire playground, the amount of sand required is equal to $\frac{1}{3}$ of the total volume of sand in bucket

For covering $1$ unit area of the entire playground, the amount of sand required is equal to

$\frac{\frac{1}{3} }{\frac{1}{9} } \\\\\frac{1}{3} * \frac{9}{1} \\\frac{9}{3}\\= 3$

For covering $1$ unit area of the entire playground, the amount of sand required is equal to volume of $3$ buckets of sand.

4 0

2 years ago

(b) A department store has 7,000 charge accounts. The comptroller takes a random sample of 36 of

galben [10]

Answer:

a.0.8664

b. 0.23753

c. 0.15866

Step-by-step explanation:

The comptroller takes a random sample of 36 of the account balances and calculates the standard deviation to be N42.00. If the actual mean (1) of the account balances is N175.00, what is the probability that the sample mean would be between

a. N164.50 and N185.50?

b. greater than N180.00?

c. less than N168.00?

We solve the above question using z score formula

z = (x-μ)/σ/√n where

x is the raw score,

μ is the population mean = N175

σ is the population standard deviation = N42

n is random number of sample = 36

a. Between N164.50 and N185.50?

For x = N 164.50

z = 164.50 - 175/42 /√36

z = -1.5

Probability value from Z-Table:

P(x = 164.50) = 0.066807

For x = N185.50

z = 185.50 - 175/42 /√36

z =1.5

Probability value from Z-Table:

P(x=185.50) = 0.93319

Hence:

P(x = 185.50) - P(x =164.50)

= 0.93319 - 0.066807

= 0.866383

Approximately = 0.8664

b. greater than N180.00?

x > N 180

Hence:

z = 180 - 175/42 /√36

z = 5/42/6

z = 5/7

= 0.71429

Probability value from Z-Table:

P(x<180) = 0.76247

P(x>180) = 1 - P(x<180) = 0.23753

c. less than N168.00?

x < N168.

z = 168 - 175/42 /√36

z = -7/42/6

z = -7/7

z = -1

Probability value from Z-Table:

P(x<168) = 0.15866

4 0

2 years ago

X-5y=33<br> 5x-6y=70<br> solve the system by the addition method

BARSIC [14]

For this case we have the following system of equations:

$x-5y = 33\\5x-6y = 70$

We multiply the first equation by -5:

$-5x + 25y = -165$

Thus, we have the equivalent system:

$-5x + 25y = -165\\5x-6y = 70$

We add the equations:

$-5x + 5x + 25y-6y = -165 + 70\\19y = -95\\y = - \frac {95} {19}\\y = -5$

We find the value of the variable "x":

$x-5 (-5) = 33\\x + 25 = 33\\x = 33-25\\x = 8$

Thus, the solution of the system is:

$(x, y) :( 8, -5)$

Answer:

$(x, y) :( 8, -5)$

7 0

3 years ago

3a. The graph shows a school's profit P for selling x lunches on one day. PART A:

Klio2033 [76]

Answer: $210

Step-by-step explanation:

i’m over here doing the same work- $210 the answer.

8 0

2 years ago