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andrew-mc [135]

3 years ago

11

in spencers garden the number of rose bushes is 7 less than 1.5 times the number of carnation bushes. If the number of carnation

bushesis c,then the expressions represents the number of rose bushes is ?

1 answer:

Step2247 [10]3 years ago

7 0

<span>1.5 times the number of carnation bushes.
If the number of carnation bushes is c.
R = 1.5c - 7
Hope this helps:-)</span>

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Let X denote the number of flaws along a 100-m reel of magnetic tape (an integer-valued variable). Suppose Zdenote the number of

Lemur [1.5K]

Answer:

a) $P(20 \leq X \leq 30) = P(20-0.5 \leq X \leq 30+0.5)$

$P(19.5 \leq X \leq 30.5) = P(\frac{19.5-25}{5} \leq Z \leq \frac{30.5 -25}{5})=P(-1.1 \leq Z \leq 1.1)$

And we can find this probability like this:

$P(-1.1 \leq Z \leq 1.1)= P(Z\leq 1.1) -P(Z\leq -1.1) = 0.864-0.136= 0.728$

b) $P(X \leq 30)= P(X$

And using the z score we got:

$P(X$

c) $P(X$

And if we use the continuity correction we got:

$P(X$

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".

Continuity correction means that we need to add and subtract 0.5 before standardizing the value specified.

Part a

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

$X \sim N(25,5)$

Where $\mu=25$ and $\sigma=5$

Part a

For this case we want to find this probability:

$P(20 \leq X \leq 30) = P(20-0.5 \leq X \leq 30+0.5)$

And if we use the z score given by:

$z =\frac{x-\mu}{\sigma}$

We got this:

$P(19.5 \leq X \leq 30.5) = P(\frac{19.5-25}{5} \leq Z \leq \frac{30.5 -25}{5})=P(-1.1 \leq Z \leq 1.1)$

And we can find this probability like this:

$P(-1.1 \leq Z \leq 1.1)= P(Z\leq 1.1) -P(Z\leq -1.1) = 0.864-0.136= 0.728$

Part b

For this case we want this probability:

$P(X \leq 30)$

And if we use the continuity correction we got:

$P(X \leq 30)= P(X$

And using the z score we got:

$P(X$

Part c

For this case we want this probability:

$P(X$

And if we use the continuity correction we got:

$P(X$

3 0

3 years ago

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Ilya [14]

Answer:

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Step-by-step explanation:

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7 0

3 years ago

5. Find the mZABC if<br> ZABD = 35°, ZDBC = 55º.

Reika [66]

Answer:

$m\angle ABC=90^o$

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

$m\angle ABC=m\angle ABD+m\angle DBC$ ----> by angle addition postulate

substitute the given values

$m\angle ABC=35^o+55^o=90^o$

5 0

3 years ago

Find an antiderivative F(x) with F′(x) = f(x) = 6 + 24x^3 + 18x^5 and F(1)=0.

7nadin3 [17]

Answer:

The antiderivative is $F(X) = 6x + 6x^4 + 3x^6 - 15$ .

Step-by-step explanation:

Antiderivative F(x)

This is the integral of $F^{\prime}(x)$

So

F′(x) = f(x) = 6 + 24x^3 + 18x^5

Then:

$F(x) = \int (6 + 24x^3 + 18x^5) dx$

$F(x) = 6x + \frac{24x^4}{4} + \frac{18x^6}{6} + K$

$F(x) = 6x + 6x^4 + 3x^6 + K$

F(1)=0

$F(X) = 0$ when $x = 1$ . We use this to find K.

$F(x) = 6x + 6x^4 + 3x^6 + K$

$0 = 6 + 6 + 3 + K$

$K = -15$

Thus

The antiderivative is $F(X) = 6x + 6x^4 + 3x^6 - 15$ .

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3 years ago

Prove the following statements: (a) For every integer x, if x is even, then for every integer y, xy is even. (b) For every integ

Murrr4er [49]

Answer and Step-by-step explanation:

(a) Given that x and y is even, we want to prove that xy is also even.

For x and y to be even, x and y have to be a multiple of 2. Let x = 2k and y = 2p where k and p are real numbers. xy = 2k x 2p = 4kp = 2(2kp). The product is a multiple of 2, this means the number is also even. Hence xy is even when x and y are even.

(b) in reality, if an odd number multiplies and odd number, the result is also an odd number. Therefore, the question is wrong. I assume they wanted to ask for the proof that the product is also odd. If that's the case, then this is the proof:

Given that x and y are odd, we want to prove that xy is odd. For x and y to be odd, they have to be multiples of 2 with 1 added to it. Therefore, suppose x = 2k + 1 and y = 2p + 1 then xy = (2k + 1)(2p + 1) = 4kp + 2k + 2p + 1 = 2(kp + k + p) + 1. Let kp + k + p = q, then we have 2q + 1 which is also odd.

(c) Given that x is odd we want to prove that 3x is also odd. Firstly, we've proven above that xy is odd if x and y are odd. 3 is an odd number and we are told that x is odd. Therefore it follows from the second proof that 3x is also odd.

3 0

3 years ago