Question 2

Math question below for question

Mashcka [7]

Answer=49

The dot in the middle represents the median

3 0

3 years ago

Read 2 more answers

If f(x)=3x-1 and g(x)=x+2,find (f+g)(x)

KiRa [710]

Answer:

(f + g)(x) = 4x + 1

Step-by-step explanation:

Given f(x) = 3x - 1 and g(x) = x + 2, find (f + g)(x):

We can rewrite (f + g)(x) as f(x) + g(x), and solve the composite functions through addition:

f(x) + g(x) = (3x - 1) + (x + 2)

Combine like terms:

f(x) + g(x) = 3x + x + 2 - 1 = 4x + 1

Therefore, (f + g)(x) = 4x + 1.

Please mark my answers as the Brainliest if you find my explanations/solution helpful :)

7 0

2 years ago

There are 30 students in Mr.Thomas' class.If he wants 20% of the students in each group, how many students will be in. each grou

zhuklara [117]

First you find out what 20% of 30 is (30 * 0.20 = 6)
Then you take 30/6 to get your answer (30/6 = 5)
There will be 5 students in each group.

8 0

3 years ago

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A metal hollow bar whose cross section and dimension are shown below weighs 8x10^3 kg/m^3 and measure 2m in length ..determine t

devlian [24]

1) We calculate the volume of a metal bar (without the hole).

volume=area of hexagon x length
area of hexagon=(3√3 Side²)/2=(3√3(60 cm)²) / 2=9353.07 cm²
9353.07 cm²=9353.07 cm²(1 m² / 10000 cm²)=0.935 m²

Volume=(0.935 m²)(2 m)=1.871 m³

2) we calculate the volume of the parallelepiped

Volume of a parallelepiped= area of the section x length
area of the section=side²=(40 cm)²=1600 cm²
1600 cm²=(1600 cm²)(1 m² / 10000 cm²=0.16 m²
Volume of a parallelepiped=(0.16 m²)(2 m)=0.32 m³

3) we calculate the volume of a metal hollow bar:
volume of a metal hollow bar=volume of a metal bar - volume of a parallelepiped

Volume of a metal hollow bar=1.871 m³ - 0.32 m³=1.551 m³

4) we calculate the mass of the metal bar

density=mass/ volume ⇒ mass=density *volume

Data:
density=8.10³ kg/m³
volume=1.551 m³

mass=(8x10³ Kg/m³ )12. * (1.551 m³)=12.408x10³ Kg

answer: The mas of the metal bar is 12.408x10³ kg or 12408 kg

4 0

2 years ago

Winona’s bowling scores for the past nine weeks are 195, 180, 195, 212, 208, 231, 179, 246, and 195. Find the mean, median, and

belka [17]

Answer: Mean = 204.6 , Median = 195 and mode =195

Step-by-step explanation:

The given data for Winona’s bowling scores for the past nine weeks : 195, 180, 195, 212, 208, 231, 179, 246, 195.

The formula to find Mean is given by :-

$\text{Mean}=\dfrac{\text{Sum of all observations}}{\text{Number of observations}}\\\\=\dfrac{1841}{9}=204.555555556\approx204.6$

For median , first we arrange data values in order as

179, 180, 195, 195, 195, 208, 212, 231, 246

Since the number of data values is 9 (odd).

∴ Median = Middle most data value = 195

Also, Mode = Most occurring data value = 195

3 0

3 years ago