All categories

3 years ago

A rucksack can hold a maximum of 6 kg.

Mathematics

2 answers:

Lemur [1.5K]3 years ago

7 0

(convert the 6kg to g first)

1kg = 1000
6kg=6000

6000 ÷ 680 = 8,82

the rucksack can hold between 8 and 9 textbooks

ratelena [41]3 years ago

5 0

Answer:

9 textbooks!

Step-by-step explanation:

Rucksack can hold = 6kg

Mass of textbook = 680g

So, here we need to show the division for finding the number of textbooks it can hold.

680 g in KG's = 0.68

Now, 6/0.68

That is 8.82 ( Round off = 9 textbooks)

You might be interested in

A ski jumper is preparing for her start on top of the jumping hill in korea which has a horizontal run of 234 meters. She sees t

slamgirl [31]

Answer:

The height of the jumping hill x = 189.3 m

Step-by-step explanation:

From Δ ABC

AB = x = height of the hill

∠A = 36° , ∠B = 90°

Thus ∠C = 180 - ∠A - ∠B

⇒ ∠C = 180 - 36 - 90

⇒ ∠C = 54°

From Δ ABC

$\sin 54 = \frac{AB}{AC}$

$\sin 54 = \frac{x}{234}$

x = 0.809 × 234

x = 189.3 m

This is the height of the jumping hill.

4 0

3 years ago

A recipe for a chocolate banana milkshake calls for 2

faltersainse [42]

Answer: 12 scoops

Step-by-step explanation:

Because 2 bananas =3

4=6

6=9

And 8

Would equal 12

4 0

3 years ago

find the smaller of 2 consecutive integers, x, if the sum of the smaller is 3 times the larger is -5. Write an equation and solv

Leno4ka [110]

You have defined x to be the smaller of two consecutive integers. Then the larger will be (x+1). The given relation is

... (x) + 3(x+1) = -5 . . . . the smaller plus 3 times the larger is -5

... 4x + 3 = -5 . . . . . . . .simplify

... 4x = -8 . . . . . . . . . . subtract 3

... x = -2 . . . . . . . . . . . divide by the coefficient of x

The smaller of the two integers is -2.

5 0

3 years ago

Which table would it be

Sladkaya [172]

Answer:

Step-by-step explanation:

7 0

3 years ago

A certain test preparation course is designed to help students improve their scores on the GRE exam. A mock exam is given at the

Westkost [7]

Answer:

(0.25 ; 21.75)

8.7 (1 decimal place)

Step-by-step explanation:

Net change in scores : X = 6,14,12,23,0

Sample mean, xbar = (6 + 14 + 12 + 23 + 0) /5 = 55 /5 = 11

Sample standard deviation, s = 8.66 ( from calculator)

Sample standard deviation = 8.7( 1 decimal place)

Sample size, n = 5

The 95% confidence interval; we use t, because n is small

Tcritical at 95%, df = 4 - 1 = 3 ; Tcritical = 2.776

Xbar ± standard Error

Standard Error = Tcritical * s/√n

Standard Error = 2.776 * 8.66/√5

Standard Error = 10.751086 = 10.75

Lower boundary = (11 - 10.75) = 0.25

Upper boundary = (11 + 10.75) = 21.75

(0.25 ; 21.75)

The

8 0

2 years ago