How much is 1 watt in volts

PLEASE HELP NEED THIS TO PASS

yaroslaw [1]

The solutions to f(x) = 64 is x = 7 and x = –7.

Solution:

Given data:

$$f(x)=x^{2}+15$ – – – – (1)

$f(x)=64$ – – – – (2)

To find the solutions to f(x) = 64.

Equate equation (1) and (2), we get

$x^2+15=64$

Subtract 15 from both sides of the equation.

$x^2+15-15=64-15$

$x^2=49$

$x^2=7^2$

Taking square root on both sides of the equation, we get

x = ±7

The solutions to f(x) = 64 is x = 7 and x = –7.

5 0

3 years ago

Solve 10/y = 5/2 The solution is y=_

kolezko [41]

Answer:

y is equal to 4.

Step-by-step explanation:

To find this, cross multiply and then divide.

10*2 = y*5

20 = 5y

4 = y

5 0

3 years ago

Rate and Ratios IF 3 kg of potatoes cost R24. How much will 7 kg of potatoes cost

balu736 [363]

Answer:

7kg of potatoes cost 56 R (dollars)

Step-by-step explanation:

First, we have to find the unit price of 1kg of potatoes. To do this we divide 24 by 3. That's 8. Each kg of potato costs 8 dollars.

Now, all we have to do is multiply 8 by 7.

8 is the unit rate/kg and 7kg is the amount they are asking.

8x7=56

7kg of potatoes cost 56R

4 0

2 years ago

Read 2 more answers

Find the volume of a HEMIsphere a radius of 4.

pentagon [3]

Answer:

honestly math is my worst subject but i think its 2 im really srry i think thats wrong tho

Step-by-step explanation:

6 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) Two computers are chosen at random.

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

LET X = Number of computers that are classified as ancient

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) Eight computers are chosen at random.

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

LET X = Number of computers that are classified as ancient

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) Here, also 8 computers are chosen at random.

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

LET X = Number of computers classified as cutting dash edge

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