Simplify (10 + 48) ÷ 2 =

Answers

Answer 1

Answer: 10+48 = 58 / 2 = 29.

hope it helps!!!!!!1

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Los datos siguientes corresponden a los tiempos de reacción de una muestra de 33 sujetos, medidos en centésimas de segundo(Inventar los dos últimos datos, reemplazando las x por un número): 55, 51, 60, 56, 64, 56, 63, 63, 61, 57, 62, 50, 49, 70, 72, 54, 48, 53, 58, 66, 68, 45, 74, 65, 58, 61, 62, 59, 64, 57, 63, xx, xx. 1. Calcular los cuartiles de los 33 sujetos 2. Calcular el decil 5 y el decil 7 3. Calcular el percentil 13 y el percentil 76

Answers

Responder:

Compruebe amablemente la explicación

Explicación paso a paso:

Dados los datos: 55, 51, 60, 56, 64, 56, 63, 63, 61, 57, 62, 50, 49, 70, 72, 54, 48, 53, 58, 66, 68, 45, 74, 65, 58, 61, 62, 59, 64, 57, 63, xx, xx

Los dos valores faltantes podrían ingresarse utilizando varios métodos, incluyendo la mediana, modal y el valor medio de los datos dados en la distribución.

Aquí usaremos el valor medio.

Media (m) de los datos = suma de la observación / número de observaciones Media =

(55 + 51 + 60 + 56 + 64 + 56 + 63 + 63 + 61 + 57 + 62 + 50 + 49 + 70 + 72 + 54 + 48 + 53 + 58 + 66 + 68 + 45 + 74 + 65 + 58 + 61 + 62 + 59 + 64 + 57 + 63) / 31 = 1844/31 = 59.4 = 59 (al número entero más cercano) Entonces reemplace los valores faltantes con 59.

En otro para hacer cálculos sobre los datos, hay que reorganizarlos.

Datos ordenados: 45,48,49,50,51,53,54,55,56,56,57,57,58,58,59,59,59,60,61,61,62,62,63,63,63, 64,64,65,66,68,70,72,74

Los cuartiles de datos:

Primer cuartil (Q1) = 1/4 (33 + 1) = 1/4 (34) = 8.5

Q1 = (8º + 9º) / 2 = (55 + 56) = 111/2 = 55,5

Segundo cuartil (Q2) = Q3 - Q1 = (25.5 - 8.5) = 17th = 59

Tercer cuartil (Q3) = 3/4 (33 + 1) = 3/4 (34) = 25.5 Q3 = (25 + 26) = (63 + 64) = 127/2 = 63.5

Rango intercuartil (IQR) = 63.5 - 55.5 = 8

Mediana = Q2 = 59

Mínimo = 45

Máximo = 74

Rango = 74-45 = 29

El quinto decil:

D5 = [5 (33 +1)] / 10 = 5 (34) / 10

D5 = 170/10 = 17º valor = 59

7mo decil: D7 = [7 (33 + 1)] / 10 = 7 (34) / 10

D7 = 238/10 = 23.8 ° término D7 = 63

Percentil:

Percentil 13 (13% × 33) = 0.13 × 33 = 4.29

4to término = 50

Percentil 76: (76% × 33) = 0.76 × 33 = 25.08

25 ° término = 63

If f(x) varies directly with x and f(x) = 2 when x = 10, then what is the value of f(x) when x = 40?a.8
b.45
c.200
d.320

Answers

Answer: a.8

Step-by-step explanation:

Given : If f(x) varies directly with x and f(x) = 2 when x = 10

We know if two quantities (x and f(x)) are directly proportional then

$(x_1)/(f(x_1))=(x_2)/(f(x_2))$

Now, for the given problem we have ,

$(10)/(2)=(40)/(f(40))\n\n\Rightarrow\ f(40)=(40*2)/(10)\n\n\Rightarrow\ f(40)=8$

Hence, if x = 40, then f(x)=8.

this is simple multiplication
do 2*4 and 10*4
so you get f(x)=8 and x=40

if you wanna do it by doing division do
2=10
f(x)=40
cross multiply then divide
so do 2*40 to get 80
then do 80/10 to get 8

Mrs. Myles gave the same test to both her first and third period class. In first period, the mean was 75 and the range was 30. In third period, the mean was 85 and the range was 50. Which is a true statement?A) Most of third period scored in the 80's.
B) More people had a score in the 70's in first period.
C) On average, third period did better than first period.
D) On average, first period did better than third period.

Answers

The right answer for the question that is being asked and shown above is that: "C) On average, third period did better than first period. " In third period, the mean was 85 and the range was 50. The statement that is true is that C) On average, third period did better than first period.

Answer:

C) On average, third period did better than first period

Step-by-step explanation:

Usa test prep

The value of a car depreciates by 18% per year. Work out the current value of a car bought 4 years ago for £20000

Answers

Given:
depreciation rate: 18%
Value of the car : 20,000
Age of the car : 4 years.

The depreciation is based on the current value of the car. Therefore, the amount of depreciation varies.

yr Beginning Value Dep. Rate Depreciation Ending Value
1   20,000.00   18% 3,600.00 16,400.00
2   16,400.00   18%      2,952.00    13,448.00
3   13,448.00   18% 2,420.64 11,027.36
4   11,027.36   18% 1,984.92 9,042.44

The current value of a car bought 4 years ago is 9,042.44

Beginning value : purchased amount on 1st year. then, ending balance of the previous year from 2nd year onwards.
Depreciation : Beginning value * depreciation rate
Ending value : Beginning value - depreciation

20000 = 100%18 x 4 = 72%20000 / 100 = 200200 = 1%200 x 72 = 14,40020000 - 14,400 = 5600The value of it now should be £5600

This is all about matrices, I don't really get them. Any help is great.

Answers

See Explanation

Step-by-step explanation:

$\begin{bmatrix} - 3 & 5 & 6\n-3 & 0 & - 3\end{bmatrix} -\begin{bmatrix} 6 & - 1& 1\n0& - 4& - 5\end{bmatrix} \n\n=\begin{bmatrix} - 3-6 & 5-(-1) & 6-1\n-3-0 & 0-(-4) & - 3-(-5)\end{bmatrix} \n\n=\begin{bmatrix} - 3-6 & 5+1 & 6-1\n-3-0 & 0+4 & - 3+5\end{bmatrix} \n\n=\begin{bmatrix} - 9 & 6 & 5\n-3 & 4 & 2\end{bmatrix} \n\n$

$\begin{bmatrix} 4 & - 6 \n-3 & - 6\n2 & 6 \end{bmatrix} + \begin{bmatrix} 5 & 3 \n-5 & 5\n 4 & - 5\end{bmatrix} \n\n= \begin{bmatrix} 4 +5& - 6+3 \n-3+(-5) & - 6+5\n2+4 & 6+(-5) \end{bmatrix} \n\n= \begin{bmatrix} 4 +5& - 6+3 \n-3-5 & - 6+5\n2+4 & 6-5 \end{bmatrix} \n\n= \begin{bmatrix} 9& - 3 \n-8 & - 1\n6 & 1\end{bmatrix} \n\n$