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3 years ago

What is the solution to -4 | -2x +6 | = -24

Mathematics

2 answers:

Norma-Jean [14]3 years ago

8 0

Answer:

please ignore my answer

Nonamiya [84]3 years ago

4 0

Answer:

0, 6 = x

Step-by-step explanation:

|-2x + 6| = 6 [Divided by -4]
Here is where you can see how to find your two x-values [first one being 0].

I hope you can see how and if this was alot of help to you, and as always, I am joyous to assist anyone at any time.

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A contractor is required by a county planning department to submit one, two, three, four, or five forms (depending on the nature

Westkost [7]

Answer:

(a) The value of k is $\frac{1}{15}$ .

(b) The probability that at most three forms are required is 0.40.

(d) $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of y.

Step-by-step explanation:

The random variable Y is defined as the number of forms required of the next applicant.

The probability mass function is defined as:

$P(y) = \left \{ {{ky};\ for \ y=1,2,...5 \atop {0};\ otherwise} \right$

(a)

The sum of all probabilities of an event is 1.

Use this law to compute the value of k.

$\sum P(y) = 1\\k+2k+3k+4k+5k=1\\15k=1\\k=\frac{1}{15}$

Thus, the value of k is $\frac{1}{15}$ .

(b)

Compute the value of P (Y ≤ 3) as follows:

$P(Y\leq 3)=P(Y=1)+P(Y=2)+P(Y=3)\\=\frac{1}{15}+\frac{2}{15}+ \frac{3}{15}\\=\frac{1+2+3}{15}\\ =\frac{6}{15} \\=0.40$

Thus, the probability that at most three forms are required is 0.40.

(c)

Compute the value of P (2 ≤ Y ≤ 4) as follows:

$P(2\leq Y\leq 4)=P(Y=2)+P(Y=3)+P(Y=4)\\=\frac{2}{15}+\frac{3}{15}+\frac{4}{15}\\ =\frac{2+3+4}{15}\\ =\frac{9}{15} \\=0.60$

Thus, the probability that between two and four forms (inclusive) are required is 0.60.

(d)

Now, for $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ to be the pmf of Y it has to satisfy the conditions:

$P(y)=\frac{y^{2}}{50}>0;\ for\ all\ values\ of\ y \\$
$\sum P(y)=1$

Check condition 1:

$y=1:\ P(y)=\frac{y^{2}}{50}=\frac{1}{50}=0.02>0\\y=2:\ P(y)=\frac{y^{2}}{50}=\frac{4}{50}=0.08>0 \\y=3:\ P(y)=\frac{y^{2}}{50}=\frac{9}{50}=0.18>0\\y=4:\ P(y)=\frac{y^{2}}{50}=\frac{16}{50}=0.32>0 \\y=5:\ P(y)=\frac{y^{2}}{50}=\frac{25}{50}=0.50>0$

Condition 1 is fulfilled.

Check condition 2:

$\sum P(y)=0.02+0.08+0.18+0.32+0.50=1.1>1$

Condition 2 is not satisfied.

Thus, $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of y.

7 0

2 years ago

Find the missing term and the missing coefficient. (6a − )5a = a2 − 35a

djverab [1.8K]

So, (6a - X)*5a = Y*(a^2) - 35a
=> 6a*5a - X*5a = Y*(a^2) - 35a 
=> 30(a^2) - X*5a = Y(a^2) - 35a

30 = Y and
X*5 = 35 or X = 7

4 0

3 years ago

The equation for a circle with center (h, k) is (x - h)^2 + (y - k)^2 = r^2. If a circle is modeled by the equation x^2 + (y + 6

Ronch [10]

(x - h)^2 + (y - k)^2 = r^2

x^2 + (y + 6)^2 = 100

we can rewrite this as

(x-0)^2 + (y - -6)^2 = 10^2

the center is (0,-6) and the radius is 10

5 0

2 years ago

Doctors are interested in seeing if lifting weights to alleviate back pain is effective. Twenty five volunteers are asked to ran

goldenfox [79]

Answer:

Comparing means from dependent samples.

Step-by-step explanation:

In the scenario described above, the 25 sampled individuals were used to test the effect of weightlifting on alleviating back pain. The samples used were not changed nor altered during the study as the same subjects who were tested to measure their level of pain before lifting weight were the same group subjected to the actual weight lifting test and again had their level of pain tested. Since only one group was involved in the study ( a single group was tested on neutral and actual treatment). Hence, the study used dependent samples to compare the average level of pain.

4 0

2 years ago

The fifth graders were given sandwiches for lunch during their field trip. Nathan ate 5/6 of his sandwich, Leroy ate 7/8 of his

riadik2000 [5.3K]

Nathan ate 20/24, Leroy ate 21/24, and Sofia ate 15/24 meaning Leroy ate the most

4 0

2 years ago

Read 2 more answers