When is -4.75 'more than' -2.25

algol13

<h3><u>Explanation</u></h3>

Given the equation.

$10 - d = - 34 - 5d$

Solve the equation for d-term.

Because we cannot subtract or add up constants and variables, we simply move the same variable term to the same side and constant term to the same constant side.

$10 + 34 = - 5d + d \\ 44 = - 4d \\ \frac{44}{ - 4} = d \\ - 11 = d$

Answer Check

Substitute d = -11 in the equation.

$10 - ( - 11) = - 34 - 5( - 11) \\ 10 + 11 = - 34 + 55 \\ 21 = 21$

The equation is true for d = -11.

<h3><u>Answer</u></h3>

<u> $\large \boxed{d = - 11}$ </u>

7 0

3 years ago

Read 2 more answers

Rob needs to mix 4.5 liters of liquid fruit punch concentrate with 0.5

Dmitriy789 [7]

4.5 litres of fruit punch with 0.5 litres of water.
13.5 /4.5 = 3. Multiply 0.5 by 3 to get 1.5 litres of water. Therefore, 13.5 litres of fruit punch concentrate needs 1.5 litres of water. 13.5 + 1.5 = 15 litres of fruit punch in total.

7 0

2 years ago

Kimberly wants to grow grass in her backyard. She is buying grass seed to plant and needs to determine the area of her backyard.

ANTONII [103]

Answer:

B) Foot

Step-by-step explanation:

6 0

2 years ago

If the cube shown above is sliced by a plane to create a rectangle that is not a square, which sets of vertices could the plane

Veronika [31]

If I’m sure it should be D or C

3 0

2 years ago

A computer can be classified as either cutting dash edge or ancient. Suppose that 94% of computers are classified as ancient.

taurus [48]

Answer:

(a) 0.8836

(b) 0.6096

(c) 0.3904

Step-by-step explanation:

We are given that a computer can be classified as either cutting dash edge or ancient. Suppose that 94% of computers are classified as ancient.

(a) <u>Two computers are chosen at random.</u>

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 2 computers

r = number of success = both 2

p = probability of success which in our question is % of computers

that are classified as ancient, i.e; 0.94

<em>LET X = Number of computers that are classified as ancient</em>

So, it means X ~ $Binom(n=2, p=0.94)$

Now, Probability that both computers are ancient is given by = P(X = 2)

P(X = 2) = $\binom{2}{2}\times 0.94^{2} \times (1-0.94)^{2-2}$

= $1 \times 0.94^{2} \times 1$

= 0.8836

(b) <u>Eight computers are chosen at random.</u>

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 8 computers

r = number of success = all 8

p = probability of success which in our question is % of computers

that are classified as ancient, i.e; 0.94

<em>LET X = Number of computers that are classified as ancient</em>

So, it means X ~ $Binom(n=8, p=0.94)$

Now, Probability that all eight computers are ancient is given by = P(X = 8)

P(X = 8) = $\binom{8}{8}\times 0.94^{8} \times (1-0.94)^{8-8}$

= $1 \times 0.94^{8} \times 1$

= 0.6096

(c) <u>Here, also 8 computers are chosen at random.</u>

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 8 computers

r = number of success = at least one

p = probability of success which is now the % of computers

that are classified as cutting dash edge, i.e; p = (1 - 0.94) = 0.06

<em>LET X = Number of computers classified as cutting dash edge</em>

So, it means X ~ $Binom(n=8, p=0.06)$

Now, Probability that at least one of eight randomly selected computers is cutting dash edge is given by = P(X $\geq$ 1)

P(X $\geq$ 1) = 1 - P(X = 0)

= $1 - \binom{8}{0}\times 0.06^{0} \times (1-0.06)^{8-0}$

= $1 - [1 \times 1 \times 0.94^{8}]$

= 1 - $0.94^{8}$ = 0.3904

Here, the probability that at least one of eight randomly selected computers is cutting dash edge is 0.3904 or 39.04%.

For any event to be unusual it's probability is very less such that of less than 5%. Since here the probability is 39.04% which is way higher than 5%.

So, it is not unusual that at least one of eight randomly selected computers is cutting dash edge.

7 0

2 years ago