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3 years ago

How many hundredths are equal to six tenths

Mathematics

1 answer:

belka [17]3 years ago

8 0

0.01x=0.6
0.01x/0.01=0.6/0.01
x=0.6/0.01 (simplify)
x=60

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Solve 5y'' + 3y' – 2y = 0, y(0) = 0, y'(0) = 2.8 y(t) = 0 Preview

mario62 [17]

Answer: The required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$5y^{\prime\prime}+3y^\prime-2y=0,~~~~~~~y(0)=0,~~y^\prime(0)=2.8~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$5m^2e^{mt}+3me^{mt}-2e^{mt}=0\\\\\Rightarrow (5m^2+3y-2)e^{mt}=0\\\\\Rightarrow 5m^2+3m-2=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow 5m^2+5m-2m-2=0\\\\\Rightarrow 5m(m+1)-2(m+1)=0\\\\\Rightarrow (m+1)(5m-1)=0\\\\\Rightarrow m+1=0,~~~~~5m-1=0\\\\\Rightarrow m=-1,~\dfrac{1}{5}.$

So, the general solution of the given equation is

$y(t)=Ae^{-t}+Be^{\frac{1}{5}t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-Ae^{-t}+\dfrac{B}{5}e^{\frac{1}{5}t}.$

According to the given conditions, we have

$y(0)=0\\\\\Rightarrow A+B=0\\\\\Rightarrow B=-A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

and

$y^\prime(0)=2.8\\\\\Rightarrow -A+\dfrac{B}{5}=2.8\\\\\Rightarrow -5A+B=14\\\\\Rightarrow -5A-A=14~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{Uisng equation (ii)}]\\\\\Rightarrow -6A=14\\\\\Rightarrow A=-\dfrac{14}{6}\\\\\Rightarrow A=-\dfrac{7}{3}.$

From equation (ii), we get

$B=\dfrac{7}{3}.$

Thus, the required solution is

$y(t)=-\dfrac{7}{3}e^{-t}+\dfrac{7}{3}e^{\frac{1}{5}t}.$

7 0

3 years ago

Suppose that a, b \in \mathbb{Z}a,b∈Z, not both 00, and let d=\gcd(a, b)d=gcd(a,b). Bezout's theorem states that dd can be writt

lara [203]

Answer:

Step-by-step explanation:

Recall that we say that d | a if there exists an integer k for which a = dk. So, let d = gcd(a,b) and let x, y be integers. Let t = ax+by.

We know that $d | a, d | b$ so there exists integers k,m such that a = kd and b = md. Then,

$t = ax+by = (kd)x+(md)y = d(kx+my)$ . Recall that since k, x, m, y are integers, then (kx+my) is also an integer. This proves that d | t.

3 0

3 years ago

Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.8. (Round your ans

Alenkinab [10]

Answer:

a) 0.011 = 1.1% probability that the sample mean hardness for a random sample of 17 pins is at least 51

b) 0.0001 = 0.1% probability that the sample mean hardness for a random sample of 45 pins is at least 51

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 normally distributed samples are 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 $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 50, \sigma = 1.8$

(a) If the distribution is normal, what is the probability that the sample mean hardness for a random sample of 17 pins is at least 51?

Here $n = 17, s = \frac{1.8}{\sqrt{17}} = 0.4366$

This probability is 1 subtracted by the pvalue of Z when X = 51. So

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

By the Central Limit Theorem

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

$Z = \frac{51 - 50}{0.4366}$

$Z = 2.29$

$Z = 2.29$ has a pvalue of 0.9890

1 - 0.989 = 0.011

0.011 = 1.1% probability that the sample mean hardness for a random sample of 17 pins is at least 51

(b) What is the (approximate) probability that the sample mean hardness for a random sample of 45 pins is at least 51?

Here $n = 17, s = \frac{1.8}{\sqrt{45}} = 0.2683$

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

$Z = \frac{51 - 50}{0.0.2683}$

$Z = 3.73$

$Z = 3.73$ has a pvalue of 0.9999

1 - 0.9999 = 0.0001

0.0001 = 0.1% probability that the sample mean hardness for a random sample of 45 pins is at least 51

8 0

3 years ago

I have an easy and quick question. So the picture shows an answered question but how do you know to divide? I just don't underst

11111nata11111 [884]

Answer:

Ok so u know to divide bc u r trying to isolate the variable and that means undoing the function in this case 0.6 is multiplied to p so u do the opposite which is division

tada

Hope this helps :)

Pls give brainliest XD :)

8 0

3 years ago

Read 2 more answers

The size of a TV is the diagonal measurement of its screen. Will a 46-inch TV fit in a rectangular cabinet measuring 32 inches l

maks197457 [2]

Answer:

sdrfffffffffsdffffffffffffffffsdffffffffffffffffffffffffffffdssss

Step-by-step explanation:

dfssssssssssssssssssssssssssssssssssssss

3 0

3 years ago