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2 years ago

(2 tens 1 one)x10 in unit form and standard form

Mathematics

1 answer:

omeli [17]2 years ago

4 0

When the number expression given as (2tens 1 one) x 10 is written in standard form, the standard form is 210 and the unit form is 2 hundred, and 1 ten

<h3>How to write the number in standard form?</h3>

The number expression is given as:

(2tens 1 one) x 10

2 tens is represented as:

2 * 10

1 one is represented as:

1 * 1

So, the number expression can be rewritten as:

(2tens 1 one) x 10 = (2 * 10 + 1 * 1) x 10

Evaluate the product

(2tens 1 one) x 10 = (20 + 1) x 10

Evaluate the sum

(2tens 1 one) x 10 = (21) x 10

Evaluate the product

(2tens 1 one) x 10 = 210

When the number expression given as (2tens 1 one) x 10 is written in standard form, the standard form is 210 and the unit form is 2 hundred, and 1 ten

Using the above steps as a guide, we have:

(5 hundreds 5 tens) * 10 ⇒ 5 thousands and 5 hundreds ⇒ 5500
(2 thousands 7 tens) / 10 ⇒ 2 hundreds and 7 units ⇒ 207
(4 ten thousands 8 hundred) / 10 ⇒ 4 thousands and 8 tens ⇒ 4080

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umka2103 [35]

Answer:

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Step-by-step explanation:

Since RY is an angle bisector then the ratio of the sides of the triangle are equal to the corresponding ratio of the base, that is

$\frac{YX}{YZ}$ = $\frac{XR}{ZR}$ , substitute values

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3 years ago

When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let

sergiy2304 [10]

Answer:

(a) P(X=3) = 0.093

(b) P(X≤3) = 0.966

(c) P(X≥4) = 0.034

(d) P(1≤X≤3) = 0.688

(e) The probability that none of the 25 boards is defective is 0.277.

(f) The expected value and standard deviation of X is 1.25 and 1.089 respectively.

Step-by-step explanation:

We are given that when circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%.

Let X = the number of defective boards in a random sample of size, n = 25

So, X ∼ Bin(25,0.05)

The probability distribution for the binomial distribution is given by;

$P(X=r)= \binom{n}{r} \times p^{r}\times (1-p)^{n-r} ; x = 0,1,2,......$

where, n = number of trials (samples) taken = 25

r = number of success

p = probability of success which in our question is percentage

of defectivs, i.e. 5%

(a) P(X = 3) = $\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$

= $2300 \times 0.05^{3}\times 0.95^{22}$

= 0.093

(b) P(X $\leq$ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= $\binom{25}{0} \times 0.05^{0}\times (1-0.05)^{25-0}+\binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$

= $1 \times 1 \times 0.95^{25}+25 \times 0.05^{1}\times 0.95^{24}+300 \times 0.05^{2}\times 0.95^{23}+2300 \times 0.05^{3}\times 0.95^{22}$

= 0.966

(c) P(X $\geq$ 4) = 1 - P(X < 4) = 1 - P(X $\leq$ 3)

= 1 - 0.966

= 0.034

(d) P(1 ≤ X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3)

= $\binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$