How to solve 700/2,5 with steps please

Need help #1. The answer is shown, but I don’t know how to get to the answer. Please teach and show steps.

Rufina [12.5K]

Answer:

B

Step-by-step explanation:

We are given that x and y are functions of time t such that x and y is a constant. So, we can write the following equation:

$x(t)+y(t)=k,\text{ where $k$ is some constant}$

The rate of change of x and the rate of change of y with respect to time t is simply dx/dt and dy/dt, respectively. So, we will differentiate both sides with respect to t:

$\displaystyle \frac{d}{dt}\Big[x(t)+y(t)\Big]=\frac{d}{dt}[k]$

Remember that the derivative of a constant is always 0. Therefore:

$\displaystyle \frac{dx}{dt}+\frac{dy}{dt}=0$

And by subtracting dy/dt from both sides, we acquire:

$\displaystyle \frac{dx}{dt}=-\frac{dy}{dt}$

Hence, our answer is B.

3 0

3 years ago

Read 2 more answers

Help can somone help please.

disa [49]

Answer:

0.7

Step-by-step explanation:

5(9n-6)

5×9n-5×6

45n-30

45n/45 -30/45

or 45n÷45-30÷45

n= 0.6666

0.7

hope dis helps!!!!!

8 0

2 years ago

Suppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 8.0 ounces and a standard deviation of 1.1

svet-max [94.6K]

Answer:

a) 0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces

b) 0.0129 = 1.29% probability that the mean weight is more than 9.0 ounces

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .