Write the numbers in a standard form: 57,900,000 227,900,000 1,427,000,000

When a constant force is applied to an object, the acceleration of the object varies inversely with its mass. When a certain con

marusya05 [52]

Answer: The mass of the object is 10 kg

Step-by-step explanation:

According to Newton's second law of motion, the force $F$ applied to an object with mass $m$ is directly proportional to its acceleration $a$ :

$F=ma$ (1)

If we isolate $a$ :

$a=\frac{F}{m}$ (2) This means the acceleration of the object varies inversely with its mass

Now, we are given the following data to calculate a constant force using (1):

$m_{1}=4 kg$

$a_{1}=5 m/s^{2}$

$F=m_{1}a_{1}$ (3)

$F=(4 kg)(5 m/s^{2)$ (4)

$F=20 N$ (5)

If we apply this same force calulated in (4) in another object with an acceleration of $a_{2}=2 m/s^{2}$ , its mass $m_{2}$ will be:

$F=m_{2}a_{2}$ (6)

$20 N=m_{2}2 m/s^{2}$ (7)

Finding $m_{2}$ :

$m_{2}=10 kg$ This is the mass of the object

4 0

3 years ago

The lifetime of a certain type of battery is normally distributed with mean value 10 hours and standard deviation 1 hour. There

Oksi-84 [34.3K]

Answer:

$z=1.64$

And if we solve for a we got;

$10 +1.64*1=11.64$

So the value of height that separates the bottom 95% of data from the top 5% is 11.64

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent certain lifetime of a population, and for this case we know the distribution for X is given by:

$X \sim N(10,1)$

Where $\mu=10$ and $\sigma=1$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.05$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.95 of the area on the left and 0.05 of the area on the right it's z=1.64. On this case P(Z<1.64)=0.95 and P(z>1.64)=0.05

If we use condition (b) from previous we have this:

$P(X$