All categories

3 years ago

A weather forecaster predicts that there is a 45% probability of snow today. What is the probability that it will not snow?

1 answer:

8 0

55% if the percent is out of 100%. in which most weather statisics are out of 100. in which you take 100-45= giving you 55%

You might be interested in

Given that (9,3) is on the graph of f(x), find the

Serjik [45]

Answer:

Step-by-step explanation:do you have the answers to choose from?

3 0

3 years ago

Read 2 more answers

write a possible point slope form of the equation of the line that passes through the points (4,4) and (3,-2)?

Mashutka [201]

m=-2-4/3-4=-6/-1=6

y-4=6(x-4)

y=6x-20

6 0

3 years ago

I dont understand I need help

gregori [183]

Answer:

7.4825 km or 7.48 km (rounded to nearest hundredth)

Step-by-step explanation:

Ranch's measurements rounded up to the nearest hundredth:

1st measurement = $\sqrt{60}$ = 7.75 km

2nd measurement = $\frac{58}{8}$ = 7.25 km

3rd measurement = 7.3(recurring) = 7.33 km

4th measurement = 7 $\frac{3}{5}$ = 7.60

The average of the four measurements is:

(7.75 + 7.25 + 7.33 + 7.60) ÷ 4 = 7.4825 km or 7.48 km (rounded to nearest hundredth)

6 0

4 years ago

The football team at Riverside College plays in an old stadium that seats 50,750 people. This stadium will be demolished and a n

Serhud [2]

69,020 please mark me brainliest

4 0

3 years ago

Read 2 more answers

Find derivative problem Find B’(6)

dalvyx [7]

Answer:

$B^\prime(6) \approx -28.17$

Step-by-step explanation:

We have:

$\displaystyle B(t)=24.6\sin(\frac{\pi t}{10})(8-t)$

And we want to find B’(6).

So, we will need to find B(t) first. To do so, we will take the derivative of both sides with respect to x. Hence:

$\displaystyle B^\prime(t)=\frac{d}{dt}[24.6\sin(\frac{\pi t}{10})(8-t)]$

We can move the constant outside:

$\displaystyle B^\prime(t)=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)]$

Now, we will utilize the product rule. The product rule is:

$(uv)^\prime=u^\prime v+u v^\prime$

We will let:

$\displaystyle u=\sin(\frac{\pi t}{10})\text{ and } \\ \\ v=8-t$

Then:

$\displaystyle u^\prime=\frac{\pi}{10}\cos(\frac{\pi t}{10})\text{ and } \\ \\ v^\prime= -1$

(The derivative of u was determined using the chain rule.)

Then it follows that:

$\displaystyle \begin{aligned} B^\prime(t)&=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)] \\ \\ &=24.6[(\frac{\pi}{10}\cos(\frac{\pi t}{10}))(8-t) - \sin(\frac{\pi t}{10})] \end{aligned}$

Therefore:

$\displaystyle B^\prime(6) =24.6[(\frac{\pi}{10}\cos(\frac{\pi (6)}{10}))(8-(6))- \sin(\frac{\pi (6)}{10})]$

By simplification:

$\displaystyle B^\prime(6)=24.6 [\frac{\pi}{10}\cos(\frac{3\pi}{5})(2)-\sin(\frac{3\pi}{5})] \approx -28.17$

So, the slope of the tangent line to the point (6, B(6)) is -28.17.

5 0

3 years ago