All categories

3 years ago

(10×2+99×-2)÷(×+10)=

Mathematics

2 answers:

hammer [34]3 years ago

5 0

the answer is: -178/x+10

svp [43]3 years ago

5 0

your answer would be, 119x - 2/ x + 10.

Thanks,

Deku ❤

You might be interested in

CAN SOMEONE HELP ME WITH MY ALGEBRA 1.

Alenkinab [10]

Answer:

-5

Step-by-step explanation:

$\frac{7-2^{2}+17}{-14+(2)(5)}$

Multiply 2 · 2 = 4

$\frac{7-4+17}{-14+(2)(5)}$

Subtract 4 - 7 = 3

$\frac{3+17}{-14+(2)(5)}$

Add 3 + 17 = 20

$\frac{20}{-14+(2)(5)}$

Multiply 2 · 5 = 10

$\frac{20}{-14+10}$

Add -14 + 10 = -4

$\frac{20}{-4}$

Divide 20 ÷ 4 = -5

$-5$

5 0

3 years ago

Read 2 more answers

Let f(x) = x2 - 5 and g(x) = 3x2. Find g(f(x))

m_a_m_a [10]

Answer:

Step-by-step explanation:

to find composite function:

put the inner funtion into the outer function

inner func.= $x^2-5$

outer= $3x^2$

hence,

$g(f(x))= 3(x^2-5)^2$

3 0

3 years ago

Which expression is equivalent to log12 (1/8/8w) ?

makkiz [27]

Hello! For this problem we need to make use of two existing logarithmic properties. These are:

1. $log \frac{M}{N}$ is equal to $log M-logN$
2. $log(MN)$ is equal to $log M+logN$

Following the first property, we can simplify the expression to $log \frac{1}{2}-log(8w)$ .

Then, we will use the second property to the term 8w. The final form of the expression would then be $log \frac{1}{2}-(log8+logw)$

ANSWER: $log \frac{1}{2}-(log8+logw)$ (second option)

5 0

4 years ago

If the starting time of a project is 11:25 and the finishing time if 13:40 , what is the duration.

antoniya [11.8K]

Answer:

The duration is 135 minutes ie. 2 hours and 15 minutes

3 0

2 years ago

Suppose that bugs are present in 1% of all computer programs. A computer de-bugging program detects an actual bug with probabili

lawyer [7]

Answer:

(i) The probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

Step-by-step explanation:

Denote the events as follows:

B = bugs are present in a computer program.

D = a de-bugging program detects the bug.

The information provided is:

$P(B) =0.01\\P(D|B)=0.99\\P(D|B^{c})=0.02$

(i)

The probability that there is a bug in the program given that the de-bugging program has detected the bug is, P (B | D).

The Bayes' theorem states that the conditional probability of an event E given that another event X has already occurred is:

$P(E|X)=\frac{P(X|E)P(E)}{P(X|E)P(E)+P(X|E^{c})P(E^{c})}$

Use the Bayes' theorem to compute the value of P (B | D) as follows:

$P(B|D)=\frac{P(D|B)P(B)}{P(D|B)P(B)+P(D|B^{c})P(B^{c})}=\frac{(0.99\times 0.01)}{(0.99\times 0.01)+(0.02\times (1-0.01))}=0.3333$

Thus, the probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii)

The probability that a bug is actually present given that the de-bugging program claims that bug is present is:

P (B|D) = 0.3333

Now it is provided that two tests are performed on the program A.

Both the test are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is:

P (Bugs are actually present | Detects on both test) = P (B|D) × P (B|D)

$=0.3333\times 0.3333\\=0.11108889\\\approx 0.1111$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii)

Now it is provided that three tests are performed on the program A.

All the three tests are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is:

P (Bugs are actually present | Detects on all 3 test)

= P (B|D) × P (B|D) × P (B|D)

$=0.3333\times 0.3333\times 0.3333\\=0.037025927037\\\approx 0.037$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

4 0

3 years ago