All categories

3 years ago

Are the rations equivalent? 2/3 =10/15?

Mathematics

1 answer:

Nesterboy [21]3 years ago

3 0

Answer:

Step-by-step explanation:

You might be interested in

Can someone help please? this is due soon. ill give brainliest

Iteru [2.4K]

Not sure but I think it’s B

4 0

3 years ago

Read 2 more answers

The weight distribution of parcels sent in a certain manner is normal with mean of 12 lb and standard deviation of 3 lb. The par

kicyunya [14]

Answer:

21.16

Step-by-step explanation:

Starting from the theory we have the following equation:

$fi*P(x$

Using the data supplied in the exercise, we have subtracting the mean and dividing by the standard deviation:

$P( z \leq \frac{c-1-12}{3.5}) =0.99/fi$

solving for "c", knowing that fi is a tabulating value:

$\frac{c-13}{3.5}=0.99/fi\\\frac{c-13}{3.5}=2.33\\c-13=2.33*3.5\\c = 8.155 +13\\c = 21.155$

therefore the value of c is equal to 21.16

8 0

4 years ago

SOMEONE PLEASEEEEE HELP .... GIVING BRAINILEST!!!

Vanyuwa [196]

Answer: Value of M: -49

Value of N: 24

Step-by-step explanation:

Yw and pls mark me brainiest

3 0

2 years ago

Read 2 more answers

Sadiq measured the dimensions of a 1 kg bag of sugar and found the length was 9.2 cm, the width was 6 cm and the height was 14.1

garik1379 [7]

Answer:

$d=1.28\times 10^{-6}\ g/cm^3$

Step-by-step explanation:

The mass of a bag of sugar, m = 1 kg = 0.001 g

Length of bag, l = 9.2 cm

Width of bag, b = 6 cm

Height of the bag, h = 14.1 cm

We need to find the density of the sugar bag.

Density = mass/volume

So,

$d=\dfrac{0.001\ g}{9.2\times 6\times 14.1\ cm^3}\\\\d=1.28\times 10^{-6}\ g/cm^3$

So, the density of the bag is $1.28\times 10^{-6}\ g/cm^3$ .

7 0

3 years ago

PLEASE HELP ME I DONT UNDERSTAND THIS AT ALL

Delicious77 [7]

What do you not understand

4 0

3 years ago