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2 years ago

5

One of the legs of a right triangle measures 3 cm and the other leg measures 14 cm. Find the measure of the hypotenuse. If neces

sary, round to the nearest tenth.

1 answer:

cupoosta [38]2 years ago

4 0

Please mark me as brainliest :)

This is *RIGHT ANGLED TRIANGLE*
c≈14.32cm

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-3(2y+4) is greater that equal to 12-4y<br>

ipn [44]

Answer:

y is less than or equal to -12

y<=-12

Step-by-step explanation:

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3 years ago

Read 2 more answers

Every valid argument with true premises has a true conclusion. Rewrite the above statement in the form V[x] x, if [y] then [z] (

just olya [345]

Answer:

Let y(x)="x is valid and x has true premises" and z(x)="x has a true conclusion".

Step-by-step explanation:

The universe U is the collection of all arguments so that x∈U. The statement uses the universal quantifier ∀ represented by the word "Every". The words "valid", "with true premises" and "has a true conclusions" are properties of an argument x.

We can interptet the statement as: "For all x, (x is valid and x has true premises)→(x has a true conclusion)". Symbolically, (∀x)(y(x)→z(x)). The implication → can be read as "if y(x) then z(x)".

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3 years ago

At the craft store, Mark bought a bag of blue and brown marbles. The bag contained 20 marbles, and 30% of them were blue. How ma

kykrilka [37]

Answer:

6 blue marbles

Step-by-step explanation:

20 marbles * .30 are blue = blue marbles

20 * .30 = 6 blue marbles

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3 years ago

The expected number of typographical errors on a page of a certain magazine is .2. What is the probability that an article of 10

Pavel [41]

Answer:

a) The probability that an article of 10 pages contains 0 typographical errors is 0.8187.

b) The probability that an article of 10 pages contains 2 or more typographical errors is 0.0175.

Step-by-step explanation:

Given : The expected number of typographical errors on a page of a certain magazine is 0.2.

To find : What is the probability that an article of 10 pages contains

(a) 0 and (b) 2 or more typographical errors?

Solution :

Applying Poisson distribution,

$N\sim Pois(0.2)$

$P(N=r)=\frac{e^{-np}(np)^r}{r!}$

where, n is the number of words in a page

and p is the probability of every word with typographical errors.

Here, n=10 and E(N)=np=0.2

a) The probability that an article of 10 pages contains 0 typographical errors.

Substitute r=0 in formula,

$P(N=0)=\frac{e^{-0.2}(0.2)^0}{0!}$

$P(N=0)=\frac{e^{-0.2}}{1}$

$P(N=0)=e^{-0.2}$

$P(N=0)=0.8187$

The probability that an article of 10 pages contains 0 typographical errors is 0.8187.

b) The probability that an article of 10 pages contains 2 or more typographical errors.

Substitute $r\geq 2$ in formula,

$P(N\geq 2)=1-P(N$

$P(N\geq 2)=1-[P(N=0)+P(N=1)]$

$P(N\geq 2)=1-[\frac{e^{-0.2}(0.2)^0}{0!}+\frac{e^{-0.2}(0.2)^1}{1!}]$

$P(N\geq 2)=1-[e^{-0.2}+e^{-0.2}(0.2)]$

$P(N\geq 2)=1-[0.8187+0.1637]$

$P(N\geq 2)=1-0.9825$

$P(N\geq 2)=0.0175$

The probability that an article of 10 pages contains 2 or more typographical errors is 0.0175.

6 0

3 years ago

1/8 (16k-24) + 1/5 (2+10k) <br><br>Show your work.

bearhunter [10]

16/8k - 24/8 + 2/5 + 10/5k

2k - 3 + 2/5 + 2k

4k - 13/5

Hope it helps :)

6 0

3 years ago