Answer:
10 terms
Step-by-step explanation:
equate the sum formula to 55 and solve for n
n(n + 1) = 55 ( multiply both sides by 2 to clear the fraction )
n(n + 1) = 110 ← distribute parenthesis on left side
n² + n = 110 ( subtract 110 from both sides )
n² + n - 110 = 0 ← in standard form
Consider the factors of the constant term (- 110) which sum to give the coefficient of the n- term (+ 1)
the factors are + 11 and - 10 , since
11 × - 10 = - 110 and 11 - 10 = + 1 , then
(n + 11)(n - 10) = 0 ← in factored form
equate each factor to zero and solve for n
n + 11 = 0 ⇒ n = - 11
n - 10 = 0 ⇒ n = 10
However, n > 0 , then n = 10
number of terms which sum to 55 is 10
Answer: 3 red marbles, 3 blue marbles, 6 yellow marbles, 8 black marbles, 4 silver marbles
Step-by-step explanation:
I know these are the correct number of marbles because I set up these numbers as fractions and solved for x.
For examples, 2/16 of the marbles are red. 2/16=x/24
You would do this because we are trying to find the number of marbles that are red out of the 24 marbles.
Next, we would do 24 divided by 16= 1.5 Then 1.5 times 2=3
You would follow this rule for each fraction of marbles in order to find the true number of marbles out of 24.
Yes, set up your proportion: x/21 = 10/14
Cross multiply 14x = 210
Divide both sides by 14 and x= 15 cm
Answer:
1.8m^2 approx
Step-by-step explanation:
Given data
P1= 1400N
A1=0.5m^2
P2=5000 N
A2=??
Let us apply the formula to calculate the Area A2
P1/A1= P2/A2
substitute
1400/0.5= 5000/A2
cross multiply
1400*A2= 5000*0.5
1400*A2= 2500
A2= 2500/1400
A2= 1.78
Hence the Area is 1.8m^2 approx
La opinión correcta es la de Camila que argumenta que la frecuencia relativa de la cara con el #6 será un valor cercano a 0,166.
<h3>Qué es la frecuencia relativa?</h3>
La frecuencia relativa es un término matemático que se utiliza en la estadística para referirse al número de veces que un evento se repite durante un experimento.
Por otra parte, es necesario aclara que la frecuencia relativa no se modifica en gran medida si se aumenta el número de veces de una prueba, debido a que la probabilidad de ocurrencia de este evento va a ser la misma.
De acuerdo a lo anterior, Camila tiene razón debido a que considera que el valor de la frecuencia relativa no se modifica en gran medida entre 300 o 600 lanzamientos de los dados.
Nota: Esta pregunta está incompleta porque no está la tabla. No obstante, la puedo responder basado en mi conocimiento previo.
Aprenda más sobre probabilidad en: brainly.com/question/16019390
