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2 years ago

Which expresion is equivalent to -5 -(-7)? HURRY 20 POINTS

Mathematics

1 answer:

grigory [225]2 years ago

3 0

Answer:

<h2>2 (the answer should have a positive sign.)</h2><h2>The equivalent to -5-(-7) is -5+7.</h2>

Step-by-step explanation:

To solve this problem, use the order of operations.

PEMDAS

Parentheses

Exponents

Multiply

Divide

Add

Subtract

$\sf{-5-(-7)=-5+7}$

$\sf{-5+7}$

Add or subtract.

Solve.

$-5+7=2$

$\large\boxed{2}$

So, the correct answer is 2.

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What could be the next line in the solution of the equation 3 (x - 2) = 15 ?

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X = 7 is the answer

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2 years ago

(1 point) Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference betw

slavikrds [6]

Answer:

a. $\frac{dT}{dt}=k(T-Tm); T(0)=190$

b. $C_{0}=122$

c. $k=-0.00259$

d. $t=153.39838\\$ minutos

Step-by-step explanation:

a. Newton's law of cooling states that the speed with which a body is cooled is proportional to the difference between its temperature and that of the medium in which it is found. Then, the initial value problem is given by:

$Tm=68$

$\frac{dT}{dt}=k(T-Tm); T(0)=190$

b. The differential equation obtained is a differential equation of separable variables:

$\frac{dT}{T-Tm}=kdt\\\\\int {\frac{dT}{T-Tm}}=\int{kdt}\\\\Ln|T-Tm|=kt+C\\\\T(t)=C_{0}e^{kt}+Tm=C_{0}e^{kt}+68\\\\T(0)=C_{0}e^{k(0)}+68=190\\\\C_{0}=122$

c. After 33 minutes of serving the coffee has cooled to 180°:

$T(33)=122e^{33k}+68=180\\\\e^{33k}=\frac{112}{122}\\\\33k=Ln(\frac{112}{122})\\\\k=-0.00259$

$150=122e^{-0.00259t}+68\\\\Ln(\frac{150-68}{122})=-0.00259t\\\\t=153.39838\\\\$

8 0

3 years ago

The sign says that the price marked on all music equipment is 30% off the regional price you buy an electric guitar for the sale

kolezko [41]

I NEED 5PT but the answer 95 and 1/2 i think

8 0

3 years ago

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This exercise illustrates that poor quality can affect schedules and costs. A manufacturing process has 90 customer orders to fi

svp [43]

Answer:

a) 0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b) 0.4062 = 40.62% probability that the 100 orders can be filled without reordering components.

c) 0.9034 = 90.34% probability that the 100 orders can be filled without reordering components

Step-by-step explanation:

For each component, there are only two possible outcomes. Either it is defective, or it is not. The components can be assumed to be independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

3% of the components are identified as defective

This means that $p = 0.03$

a. If the manufacturer stocks 90 components, what is the probability that the 90 orders can be filled without reordering components?

0 defective in a set of 90, which is $P(X = 0)$ when $n = 90$ . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{90,0}.(0.03)^{0}.(0.97)^{90} = 0.0645$

0.0645 = 6.45% probability that the 90 orders can be filled without reordering components.

b. If the manufacturer stocks 102 components, what is the probability that the 100 orders can be filled without reordering components?

At most 102 - 100 = 2 defective in a set of 102, so $P(X \leq 2)$ when $n = 102$

Then

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{102,0}.(0.03)^{0}.(0.97)^{102} = 0.0447$

$P(X = 1) = C_{102,0}.(0.03)^{1}.(0.97)^{101} = 0.1411$

$P(X = 2) = C_{102,2}.(0.03)^{2}.(0.97)^{100} = 0.2204$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0447 + 0.1411 + 0.2204 = 0.4062$

0.4062 = 40.62% probability that the 100 orders can be filled without reordering components.

c. If the manufacturer stocks 105 components, what is the probability that the 100 orders can be filled without reordering components?

At most 105 - 100 = 5 defective in a set of 105, so $P(X \leq 5)$ when $n = 105$

Then

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{105,0}.(0.03)^{0}.(0.97)^{105} = 0.0408$

$P(X = 1) = C_{105,0}.(0.03)^{1}.(0.97)^{104} = 0.1326$

$P(X = 2) = C_{105,2}.(0.03)^{2}.(0.97)^{103} = 0.2133$

$P(X = 3) = C_{105,3}.(0.03)^{3}.(0.97)^{102} = 0.2265$

$P(X = 4) = C_{105,4}.(0.03)^{4}.(0.97)^{101} = 0.1786$

$P(X = 5) = C_{105,5}.(0.03)^{5}.(0.97)^{100} = 0.1116$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0408 + 0.1326 + 0.2133 + 0.2265 + 0.1786 + 0.1116 = 0.9034$

0.9034 = 90.34% probability that the 100 orders can be filled without reordering components

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3 years ago

A circle has its center at the origin, and (5, -12) is a point on the circle. How long is the radius of the circle?

kolezko [41]

(x-center)^2+(y-center)^2=r^2
x^2+y^2=r^2
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25+144=r^2
169=r^2
r=13

5 0

3 years ago

Read 2 more answers