Can someone plz help me with this hint it’s not D plz someone help me I’m not sure I understand this

Need help asap!!!!!!!!! :) will mark brainliest if correct!!!! -4 1/3+(-3/5)

Paha777 [63]

Answer:

Step-by-step explanation:

you do the ( ) first then you do plus and get the answer

5 0

3 years ago

It takes 10 people 5 days to build 25 machines. How many days will it take 15 people to build 50 machines?

Ray Of Light [21]

Answer:

on my thing i put 7 1/2 but the answer is 6 2/3

5 0

4 years ago

Lin solved the equation

Alekssandra [29.7K]

Answer: x=1/2

she was wrong, it's supposed to be

Step-by-step explanation: 8(x−3)+7=2x(4−17)

Step 1: Simplify both sides of the equation.

8(x−3)+7=2x(4−17)

(8)(x)+(8)(−3)+7=2x(4−17)(Distribute)

8x+−24+7=−26x

(8x)+(−24+7)=−26x(Combine Like Terms)

8x+−17=−26x

8x−17=−26x

Step 2: Add 26x to both sides.

8x−17+26x=−26x+26x

34x−17=0

Step 3: Add 17 to both sides.

34x−17+17=0+17

34x=17

Step 4: Divide both sides by 34.

34x/34=17/34 x=1/2

3 0

3 years ago

vDetermine if a Poisson experiment is described, and select the best answer: Suppose we knew that the average number of typos in

goldenfox [79]

Answer:

probability that a randomly selected page that contains only text will contain no typos that is

P(x=0) = $e^{-0.08}$ = 0.923

Step-by-step explanation:

Poisson distribution:-

Explanation of the Poisson distribution :-

The Poisson distribution can be derived as a limiting case of the binomial

distribution under the conditions that

i) p is very small

ii) n is very large

ii) λ = np (say finite

The probability of 'r' successes = $\frac{e^{-\alpha }\alpha^r }{r!}$

Given the average number of typos ∝ = 0.08 per page.

probability that a randomly selected page that contains only text will contain no typos that is = $p(x=0) = \frac{e^{-0.08 }\(-0.08)^0 }{0!}$

After calculation P(x=0) = $e^{-0.08}$ = 0.923

probability that a randomly selected page that contains only text will contain no typos =0.923

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:

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

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

Hence, the amount of oil was decreasing at 69300 barrels, yearly

7 0

3 years ago