Ask question

Ask question

All categories

3 years ago

15

translate the sentence into an equation. eight more than the product of a number and 2 is equal to 4. use the variable b for the

unknown number.'

1 answer:

Korolek [52]3 years ago

5 0

6 + 8x, where x is a number. You can also write this in standard form as 8x + 6.

You might be interested in

Apply the distributive property to factor out the greatest common factor of 12+20

Lunna [17]

4(3+5) --> this means that 12/4=3 and 20/4=5 and 4 is the GCF

3 0

3 years ago

How many millimetrs are there in 4cm?

blagie [28]

Since 1 cm = 10 mm
Then 4 cm = 40 mm

6 0

4 years ago

What is it help pleaseee

timurjin [86]

Answer:

25.13cm

Step-by-step explanation:

formula for circumference of a circle is (2r)π with r being the radius

since the problem gave us the diameter already we can change the formula to 8π (2r is equal to the diameter since diameter is twice as long as the radius)

this gives us the circumference of 25.13

6 0

2 years ago

Vincent has set a financial goal to purchase a home in five years. He wants a two-bedroom house with a small backyard in the nor

xz_007 [3.2K]

Step-by-step explanation:

In Figure 5.9 below

ABCD is a parallelogram

with m(∠ABC) = 43o

CD

. A

line through A meets

at E and m(∠AED) = 68o

a. m(∠ADE)

.

Find

b. m(∠ DAE)

c. m(∠BCD)

6 0

2 years ago

Sali throws an ordinary fair 6 sided dice once.

xenn [34]

a) 1/6

b) 1/36

c)

1H, 2H, 3H, 4H, 5H, 6H

1T, 2T, 3T, 4T, 5T, 6T

Step-by-step explanation:

a)

The probability of a certain event A to occur is given by

$p(A)=\frac{a}{n}$

where

a is the number of successfull outcomes, in which event A occurs

n is the total number of possible outcomes

In this problem, the event is

"getting a 6 when throwing a dice once"

We know that the possible outcomes of a dice are six: 1, 2, 3, 4, 5, 6, so we he have

$n=6$

The successfull outcome in this case is only if we get a 6, so only 1 outcome, therefore

$a=1$

So, the probability of this event is

$p(6)=\frac{1}{6}$

b)

In this case instead, we are throwing the dice twice.

The two throws of the dice are independent events (one does not depend on the other): the probability that two independent events A and B occur at the same time is given by the product of the individual probabilities,

$p(AB)=p(A)\cdot p(B)$

where

$p(A)$ is the probability that event A occurs

$p(B)$ is the probability that event B occurs

Here we have:

- Event A is "getting a 6 in the first throw of the dice". We already calculated this probability in part a), and it is

$p(A)=\frac{1}{6}$

- Event B is "getting a 6 in the second throw of the dice". Since the dice has not changed, the probability is still the same, so

$p(B)=\frac{1}{6}$

Therefore, the probability of getting a 6 on both throws is:

$p(66)=p(6)\cdot p(6)=\frac{1}{6}\cdot \frac{1}{6}=\frac{1}{36}$

c)

In this problem, we have:

- A dice that is thrown once

- A coin that is also thrown once

The dice has 6 possible outcomes, as we stated in part a):

1, 2, 3, 4, 5, 6

While the coin has two possible outcomes:

H = head

T = tail

So, in order to find all the outcomes of the two events combined, we have to combine all the outcomes of the dice with all the outcomes of the coin.

Doing so, using the following notation:

1H (getting 1 with the dice, and head with the coin)

The possible outcomes are:

1H, 2H, 3H, 4H, 5H, 6H

1T, 2T, 3T, 4T, 5T, 6T

So, we have a total of 12 possible outcomes.

4 0

3 years ago