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3 years ago

Solve for z -p(d+z) = -2z+59−p(d+z)=−2z+59

1 answer:

6 0

$-p(d+z)=-2z+59\qquad\text{use distributive property}\\\\-pd-pz=-2z+59\qquad\text{add pd to both sides}\\\\-pz=-2z+pd+59\qquad\text{add 2z to both sides}\\\\2z-pz=pd+59\\\\z(2-p)=pd+59\qquad\text{divide both sides by}\ (2-p)\neq0\\\\\boxed{z=\dfrac{pd+59}{2-p}}$

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Can someone pleaseeee help and if you’re correct i’ll give brainliest

deff fn [24]

Answer:

The answer should be 13.

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3 years ago

Read 2 more answers

Two planes A and B start from the same place and move in different directions making an angle of 50° between them the speed of p

Alika [10]

Answer:

The distance between them is 230.65 miles

Step-by-step explanation:

Here we use the Cosine formula

$a^2 = b^2 + c^2 - 2bc\ cos\ A\\\\a^2 = 200^2 + 300^2 - 2\times 200\times 300cos50^{\circ}\\\\= 40,000 + 90,000 - 120,000cos50^{\circ}\\\\= 130,000 - 120,000 \times 0.64\\\\= 130,000 - 76,800\\\\a^2= 53,200\\\\a = \sqrt{53,200} \\\\= 230.65\ miles\ per\ hour$

Now the distance for one hour is

= 230.65 ÷ 1

= 230.65 miles

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3 years ago

Divya starts a painting her grandpa's kitchen at 8:15 a.m. She finishes 75 minutes later

Mila [183]

Answer:

9:30 she finishes

Step-by-step explanation:

815 plus 75 is 890

there are 60 minutes in one hour

890-60 is 830

add 100 to represent another Hour

830 plus 100 is 930

9:30

therefore, it is 9:30

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3 years ago

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.

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3 years ago

What is the answer to x^6/x^4

Natalka [10]

X^2
You subtract exponents when dividing exponents

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3 years ago