Is 0.21 rational or irrational

doors for the small cabinet are 12.5 inches long. Doors for the large cabinet are 2.5 times as long as the doors from the small

Firlakuza [10]

Large cab doors= 12.5x2.5= 31.25in

1feet = 12in

12feet X 12 = 144in board

144 / 31.25 = 4.608

4 large doors can be cut

19in left over

4 0

3 years ago

The difference of two trinomials is x2 − 10x + 2. If one of the trinomials is 3x2 − 11x − 4, then which expression could be the

12345 [234]

Answer:

$4x^2-21x-2$

Step-by-step explanation:

Given that:

Difference of two trinomials is $x^2 - 10x + 2$

One of the two trinomials is $3x^2 - 11x - 4$

To find:

The other trinomial = ?

Four options are:

$2x2 - x - 2 \\2x2 + x + 6 \\4x2 + 21x + 6\\ 4x2 - 21x - 2$

Solution:

Let the two trinomials be A and B.

Given A - B = $x^2 - 10x + 2$

B = $3x^2 - 11x - 4$

We have to find the other trinomial A.

A - B = $x^2 - 10x + 2$

A - ( $3x^2 - 11x - 4$ ) = $x^2 - 10x + 2$

$\Rightarrow$ A = $x^2 - 10x + 2$ + ( $3x^2 - 11x - 4$ )

$\Rightarrow$ A = $4x^2-21x-2$

So, the correct answer is $4x^2-21x-2$ .

3 0

3 years ago

Given that f(x) = x² + 3x 28 and g(x) = x + 7, find f(x) + g(x) and

ozzi

Answer:

f(x) + g(x) = x^2 + 4x + 35

Step-by-step explanation:

Add like terms from f(x) and g(x):

f(x) + g(x) = x^2 + 3x + 28 + x + 7, or

f(x) + g(x) = x^2 + 4x + 35

7 0

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.

(c) The probability that between two and four forms (inclusive) are required is 0.60.

(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

3 years ago

Pleaseeeee answer I will mark brainliest for best answer ^>^

Orlov [11]

Answer: 7.6

Explanation:

8 0

3 years ago

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