All categories

3 years ago

A weather forecaster says the temperature will be about -5 Celsius “give or take” 10 degrees.

2 answers:

5 0

The weather forecaster means -5 +/- 10
The greatest possible temperature is 5 Celsius
The least possible temperature is -15 Celsius

svet-max [94.6K]3 years ago

4 0

Answer:

Greatest temperature would be 5 degrees celsius

Least possible temperature would be -15 degrees celsius

Step-by-step explanation:

When we say "give" it means adding the value

So giving 10 degrees would mean adding 10 to -5 degrees

-5+10 = 5 degrees celsius

When we say "take" it means subtracting the value

So take 10 degrees would mean subtracting 10 from -5 degrees

-5-10 = -15 degrees celsius

Greatest temperature would be 5 degrees celsius

Least possible temperature would be -15 degrees celsius

You might be interested in

Consider the following equation. f(x, y) = y3/x, P(1, 2), u = 1 3 2i + 5 j (a) Find the gradient of f. ∇f(x, y) = Correct: Your

BaLLatris [955]

$f(x,y)=\dfrac{y^3}x$

a. The gradient is

$\nabla f(x,y)=\dfrac{\partial f}{\partial x}\,\vec\imath+\dfrac{\partial f}{\partial y}\,\vec\jmath$

$\boxed{\nabla f(x,y)=-\dfrac{y^3}{x^2}\,\vec\imath+\dfrac{3y^2}x\,\vec\jmath}$

b. The gradient at point P(1, 2) is

$\boxed{\nabla f(1,2)=-8\,\vec\imath+12\,\vec\jmath}$

c. The derivative of $f$ at P in the direction of $\vec u$ is

$D_{\vec u}f(1,2)=\nabla f(1,2)\cdot\dfrac{\vec u}{\|\vec u\|}$

It looks like

$\vec u=\dfrac{13}2\,\vec\imath+5\,\vec\jmath$

so that

$\|\vec u\|=\sqrt{\left(\dfrac{13}2\right)^2+5^2}=\dfrac{\sqrt{269}}2$

Then

$D_{\vec u}f(1,2)=\dfrac{\left(-8\,\vec\imath+12\,\vec\jmath\right)\cdot\left(\frac{13}2\,\vec\imath+5\,\vec\jmath\right)}{\frac{\sqrt{269}}2}$

$\boxed{D_{\vec u}f(1,2)=\dfrac{16}{\sqrt{269}}}$

7 0

3 years ago

Can anyone solve the last question?

sesenic [268]

As stated in (i), the equation of the line is: ln y = -0.015x + .26
(By the way, I checked your answers for parts (i) and (ii) and they are both correct)

(iii)
Plug in (1.1) for y and solve:
ln (1.1) = -0.015x + .26
0.095 = -0.015x + .26
-0.165 = - 0.015x
10.979 = x

Answer: x = 10.979

7 0

3 years ago

A certain plant runs three shifts per day. Of all the items produced by the plant, 50% of them are produced on the first shift,

Snezhnost [94]

Answer:

0.2941 = 29.41% probability that it was manufactured during the first shift.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Defective

Event B: Manufactured during the first shift.

Probability of a defective item:

1% of 50%(first shift)

2% of 30%(second shift)

3% of 20%(third shift).

$P(A) = 0.01*0.5 + 0.02*0.3 + 0.03*0.2 = 0.017$

Probability of a defective item being produced on the first shift:

1% of 50%. So

$P(A \cap B) = 0.01*0.5 = 0.005$

What is the probability that it was manufactured during the first shift?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.005}{0.017} = 0.2941$

0.2941 = 29.41% probability that it was manufactured during the first shift.

6 0

3 years ago

The random variable X is exponentially distributed, where X represents the waiting time to be seated at a restaurant during the

erastova [34]

Answer:

The probability that the wait time is greater than 14 minutes is 0.4786.

Step-by-step explanation:

The random variable X is defined as the waiting time to be seated at a restaurant during the evening.

The average waiting time is, β = 19 minutes.

The random variable X follows an Exponential distribution with parameter $\lambda=\frac{1}{\beta}=\frac{1}{19}$ .

The probability distribution function of X is:

$f(x)=\lambda e^{-\lambda x};\ x=0,1,2,3...$

Compute the value of the event (X > 14) as follows:

$P(X>14)=\int\limits^{\infty}_{14} {\lambda e^{-\lambda x}} \, dx=\lambda \int\limits^{\infty}_{14} {e^{-\lambda x}} \, dx\\=\lambda |\frac{e^{-\lambda x}}{-\lambda}|^{\infty}_{14}=e^{-\frac{1}{19} \times14}-0\\=0.4786$

Thus, the probability that the wait time is greater than 14 minutes is 0.4786.

7 0

3 years ago

Pls answer this with an explanation if you can pg4 pt1

lesya [120]

I don't know the official answer but I can confirm for you that it is either C or D. I just don't remember how to determine which way the sign goes. Sorry

5 0

3 years ago

Read 2 more answers