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1 year ago

7

The area of a rhombus is 72 m2. One diagonal is 18m. Find the other diagonal.

1 answer:

miskamm [114]1 year ago

7 0

Answer:

Other diagonal is 8 m

Step-by-step explanation:

area of rhombus = 72 m²

½ × diagonal 1 × diagonal 2 = 72

½ × 18 × d2 = 72

9 × d2 = 72

d2 = 72/9

d2 = 8

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32% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a

Tems11 [23]

Answer:

a) $P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211$

b) $P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668$

c) $P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816$

Step-by-step explanation:

1) Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

2) Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=10, p=0.32)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

$P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211$

Part b

$P(X> 2)=1-P(X\leq 2)=1-[P(X=0)+P(X=1)+P(X=2)]$

$P(X=0)=(10C0)(0.32)^0 (1-0.32)^{10-0}=0.0211$

$P(X=1)=(10C1)(0.32)^1 (1-0.32)^{10-1}=0.0995$

$P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211$

$P(X> 2)=1-P(X\leq 2)=1-[0.0211+0.0995+0.211]=0.668$

Part c

$P(2 \leq x \leq 5)=P(X=2)+P(X=3)+P(X=4)+P(X=5)$

$P(X=2)=(10C2)(0.32)^2 (1-0.32)^{10-2}=0.211$

$P(X=3)=(10C3)(0.32)^3 (1-0.32)^{10-3}=0.264$

$P(X=4)=(10C4)(0.32)^4 (1-0.32)^{10-4}=0.218$

$P(X=5)=(10C5)(0.32)^5 (1-0.32)^{10-5}=0.123$

$P(2 \leq x \leq 5)=0.211+0.264+0.218+0.123=0.816$

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Step-by-step explanation:

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Missing term = –2xy

Solution:

Let us first find the quotient of $-8x^2y^3 \div xy$ .

$-8x^2y^3 \div xy=\frac{-8x^2y^3 }{xy}$

$=\frac{-8\times x\times x\times y\times y\times y}{xy}$

Taking common term xy outside in the numerator.

$=\frac{xy(-8\times x\times y\times y)}{xy}$

Both xy in the numerator and denominator are cancelled.

$=-8xy^2$

Thus, the quotient of $-8x^2y^3 \div xy$ is $-8xy^2$ .

Given the quotient of $-8x^2y^3 \div xy$ is same as the product of 4xy and ____.

$-8xy^2=4xy$ × missing term

Divide both sides by 4xy, we get

⇒ missing term = $\frac{-8xy^2}{4xy}$

Cancel the common terms in both numerator and denominator.

⇒ missing term = –2xy

Hence the missing term of the product is –2xy.

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