All categories

2 years ago

Expand the following 3(x+9)

Mathematics

1 answer:

mina [271]2 years ago

5 0

Answer: I'm pretty sure its 3x+27

Step-by-step explanation: So all you have to do is multiply everything in the parentheses by 3. So you would do 3 times x and it would be 3x and then 3 times 9 which will get you 27. After you just put it back into a equation 3x+27

You might be interested in

Change 3/2 to a mixed number

Mariulka [41]

Okay first you have to divide
2 goes into 3
3/2
2 fits into 3 once
1 and the left over is 1 so
1 1/2

4 0

3 years ago

A. The lengths of pregnancies in a small rural village are normally distributed with a mean of 266 days and a standard deviation

Darya [45]

Step-by-step explanation:

mean = 266

sd = 14

cumulative probability = 0.01 so the standard score = -2.33 and 2.33 to the right and left

we find X-upper and X-lower

X-lower = 266-2.33*14 = 233.38

X-upper = 266+2.33*14 = 298.62

Between 233.38 and 298.62

we have sample size = 35

X-lower = 266-2.33*14/√35 = 260.49

X-upper = 266+2.33*14/√35 = 271.5

Between 260.49 and 271.5

b. cumulative probaility = 0.25

standard score = 1.96 to the right and left

x-lower = 6.9-1.96x0.9 = 5.14

x-upper = 6.9+1.96x0.9 = 8.66

Between 5.14 and 8.66

if sample size = 45

x-lower = 6.9-1.96*0.9/√45 = 6.64

x-upper = 6.9+1.96*0.9/√45 = 7.2

Between 6.64 and 7.2

c. standard scores would have cut off value at 0.67 and -0,67

x-lower = 265.3-0.67x15.2 = 255.12

x-upper = 265.3+0.67x15.2 = 275.48

Between 255.12 and 275.48

d. we will have critical values at 1.00 and -1.00

X-lower = 265-1x16 = 249

x-upper = 265+1x16 = 281

Between 249 and 281

with sample size = 44

x-lower = 265-1x16/√44 = 262.59

x-upper = 265+1x16/√44 = 267.41

Between 262.59 and 267.41

3 0

2 years ago

Ethan decided to give 10% of his monthly income to charity. This month, he wrote the

Blizzard [7]

put the decimal in the multiplication problem in the exact same point under the line then multiply as normal and you should get 152.6 meaning $152.60

5 0

3 years ago

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were millio

JulijaS [17]

Answer:

The amount of oil was decreasing at 69300 barrels, yearly

Step-by-step explanation:

Given

$Initial =1\ million$

$6\ years\ later = 500,000$

Required

At what rate did oil decrease when 600000 barrels remain

To do this, we make use of the following notations

t = Time

A = Amount left in the well

So:

$\frac{dA}{dt} = kA$

Where k represents the constant of proportionality

$\frac{dA}{dt} = kA$

Multiply both sides by dt/A

$\frac{dA}{dt} * \frac{dt}{A} = kA * \frac{dt}{A}$

$\frac{dA}{A} = k\ dt$

Integrate both sides

$\int\ {\frac{dA}{A} = \int\ {k\ dt}$

$ln\ A = kt + lnC$

Make A, the subject

$A = Ce^{kt}$

$t = 0\ when\ A =1\ million$ i.e. At initial

So, we have:

$A = Ce^{kt}$

$1000000 = Ce^{k*0}$

$1000000 = Ce^{0}$

$1000000 = C*1$

$1000000 = C$

$C =1000000$

Substitute $C =1000000$ in $A = Ce^{kt}$

$A = 1000000e^{kt}$

To solve for k;

$6\ years\ later = 500,000$

i.e.

$t = 6\ A = 500000$

So:

$500000= 1000000e^{k*6}$

Divide both sides by 1000000

$0.5= e^{k*6}$

Take natural logarithm (ln) of both sides

$ln(0.5) = ln(e^{k*6})$

$ln(0.5) = k*6$

Solve for k

$k = \frac{ln(0.5)}{6}$

$k = \frac{-0.693}{6}$

$k = -0.1155$

Recall that:

$\frac{dA}{dt} = kA$

Where

$\frac{dA}{dt}$ = Rate

So, when

$A = 600000$

The rate is:

$\frac{dA}{dt} = -0.1155 * 600000$

$\frac{dA}{dt} = -69300$

<em>Hence, the amount of oil was decreasing at 69300 barrels, yearly</em>

7 0

2 years ago

−2(x+1)=x−3x−2 No solution, one solution, or infinite solutions

VLD [36.1K]

Answer:

x=3

Step-by-step explanation:

Both provide ATP molecules, but only fermentation occurs when oxygen is present.

3 0

2 years ago

Read 2 more answers