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2 years ago

14

State whether the following statements are true or false mathematics.

exFormula1" title=" \geqslant" alt=" \geqslant" align="absmiddle" class="latex-formula">

2 answers:

Bingel [31]2 years ago

8 0

Answer:

Statements? Your question is incomplete.

dangina [55]2 years ago

4 0

Bro which statements?

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B. Solve the following Polynomials:<br>1. u³ + 64

Art [367]

Answer:

(4 + u)(16 – 4u + u2)

Step-by-step explanation:

64 + u3.

= 43 + u3.

= (4 + u)(42 – 4.u + u2).

= (4 + u)(16 – 4u + u2).

3 0

3 years ago

4. Seleccione la respuesta correcta. Select the right answi

WITCHER [35]

Answer:

B

Step-by-step explanation:

The answer is B I have seen this question befor

5 0

3 years ago

D=6, Dx=-6, Dy=-24, Dz=-30 what is the solution set

Elis [28]

Answer:

(x, y, z) = (-1, -4, -5)

Step-by-step explanation:

(x, y, z) = (Dx, Dy, Dz)/D = (-6, -24, -30)/6

(x, y, z) = (-1, -4, -5)

3 0

3 years ago

There are many cylinders with a height of 9 inches. Let r represent the radius in

mylen [45]

Answer:

Step-by-step explanation:

V, r, h are the volume, radius, and height, respectively.

V = πr²h = 9πr²

The only error in your previous answers was that you forgot the π.

if r = 1, V = 9π1² = 9π ≅ 28 in³

if r = 2, V = 9π2² = 36π ≅ 113 in³

if r = 3, V = 9π3² = 81π ≅ 254 in³

8 0

3 years ago

A statistics textbook chapter contains 60 exercises, 6 of which are essay questions. A student is assigned 10 problems. (a) What

Scilla [17]

Answer:

a) P=0.3174

b) P=0.4232

c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

$P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}$

a) What is the probability that none of the questions are essay?

$P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174$

b) What is the probability that at least one is essay?

$P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232$

c) What is the probability that two or more are essay?

$P(X\geq2)=1-(P(0)+P(1))=1-(0.3174+0.4232)=1-0.7406=0.2594$

8 0

3 years ago