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Marta_Voda [28]
4 years ago
13

Solve the following inequality for t. 12.3+t<5.6

Mathematics
2 answers:
luda_lava [24]4 years ago
4 0
The answer in inequality form is t< -6.7
saveliy_v [14]4 years ago
4 0

Answer:

Step-by-step explanation:

t< 6.7

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Solve the following problems : Given: S, T, and U are the midpoints of RP , PQ , and QR respectively. Prove: △SPT≅△UTQ.
AleksandrR [38]

Answer:

Hence Proved △ SPT ≅ △ UTQ

Step-by-step explanation:

Given: S, T, and U are the midpoints of Segment RP , segment PQ , and segment QR respectively of Δ PQR.

To prove: △ SPT ≅ △ UTQ

Proof:

∵ T is is the midpoint of PQ.

Hence PT = PQ    ⇒equation 1

Now,Midpoint theorem is given below;

The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.

By, Midpoint theorem;

TS║QR

Also, TS = \frac{1}{2} QR

Hence, TS = QU (U is the midpoint QR) ⇒ equation 2

Also by Midpoint theorem;

TU║PR

Also, TU = \frac{1}{2} PR

Hence, TU = PS (S is the midpoint QR) ⇒ equation 3

Now in △SPT and △UTQ.

PT = PQ (from equation 1)

TS = QU (from equation 2)

PS = TU (from equation 3)

By S.S.S Congruence Property,

△ SPT ≅ △ UTQ ...... Hence Proved

6 0
3 years ago
Brian’s school locker has a three-digit combination lock that can be set using the numbers 5 to 9 (including 5 and 9), without r
andreev551 [17]
The possible digits are: 5, 6, 7, 8 and 9. Let's mark the case when the locker code begins with a prime number as A and the case when <span>the locker code is an odd number as B. We have 5 different digits in total, 2 of which are prime (5 and 7).

First propability:
</span>P_A=\frac{2}{5}=40\%
<span>
By knowing that digits don't repeat we can say that code is an odd number in case it ends with 5, 7 or 9 (three of five digits).

Second probability:
</span>P_B=\frac{3}{5}=60\%
5 0
3 years ago
En la final de un concurso escolar de
Sati [7]

Usando la distribución hipergeométrica,  la probabilidad de que los alumnos del colegio A obtengan los dos premios (no hay empates), es: 0.2 = 20%.

<h3>¿Qué es la fórmula de distribución hipergeométrica?</h3>

La fórmula es:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

C_{n,x} = \frac{n!}{x!(n-x)!}

Los parámetros son:

  • x es el número de éxitos.
  • N es el tamaño de la población.
  • n es el tamaño de la muestra.
  • k es el número total de resultados deseados.

Los valores de los parámetros son:

N = 6, k = 3, n = 2.

La probabilidad de que los alumnos del colegio A obtengan los dos premios (no hay empates), es P(X = 2), por eso:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

P(X = 2) = h(2,6,2,3) = \frac{C_{3,2}C_{3,0}}{C_{6,2}} = 0.2

Puede-se aprender más a cerca de la distribución hipergeométrica en brainly.com/question/24826394

#SPJ1

4 0
2 years ago
1/4+1/2 find the sum using number line
Arte-miy333 [17]
It's 3/4

no number line provided for me, but if you find the common denominator (4), and convert 1/2 into 2/4, then adding 2/4+1/4 is easy, and equals 3/4
3 0
4 years ago
84 divided by a number z
Xelga [282]
84/z is your answer.
8 0
4 years ago
Read 2 more answers
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