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Slav-nsk [51]
2 years ago
14

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

cm.
Calculate the density of the sugar in g/cm3
Mathematics
1 answer:
garik1379 [7]2 years ago
7 0

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.

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Evaluate 2d+3 when d =8
Snezhnost [94]

Answer:

19

Step-by-step explanation:

2(8) + 3

16 + 3

19

7 0
3 years ago
What is the improper fraction for 8 and 3/47​
erastovalidia [21]

Answer:

379/47

Step-by-step explanation:

1.  Simplify

2. Reduce fraction to lowest terms

1 is the greatest common divisor of 3 and 47. The result can't be further reduced.

3.  Convert mixed number to improper fraction

4 0
2 years ago
Read 2 more answers
Your test scores for the semester are 87,84 and 85.Can you raise your test average to 90 with you next test?
Rainbow [258]

Answer:

Yes

Step-by-step explanation:

If you add 87.84, 85, and 100 you get 272.84. Once you divide that by 3 you get 90.946, which would be your test average if you score 100 on your next test.

6 0
2 years ago
A random variable x follows a normal distribution with mean d and standard deviation o=2. It is known that x is less than 5 abou
Vaselesa [24]

Answer:

The mean of this distribution is approximately 3.96.

Step-by-step explanation:

Here's how to solve this problem using a normal distribution table.

Let z be the

\displaystyle z = \frac{x - \mu}{\sigma}.

In this question, x = 5 and \sigma = 2. The equation becomes

\displaystyle z = \frac{5 - \mu}{2}.

To solve for \mu, the mean of this distribution, the only thing that needs to be found is the value of z. Since

The problem stated that P(X \le 5) = 69.85\% = 0.6985. Hence, P(Z \le z) = 0.6985.

The problem is that the normal distribution tables list only the value of P(0 \le Z \le z) for z \ge 0. To estimate  z from P(Z \le z) = 0.6985, it would be necessary to find the appropriate

Since P(Z \le z) = 0.6985 and is greater than P(Z \le 0) = 0.50, z > 0. As a result, P(Z \le z) can be written as the sum of P(Z < 0) and P(0 \le Z \le z). Besides, P(Z < 0) = P(Z \le 0) = 0.50. As a result:

\begin{aligned}&P(Z \le z)\\ &= P(Z < 0) + P(0 \le Z \le z) \\ &= 0.50 + P(0 \le Z \le z)\end{aligned}.

Therefore:

\begin{aligned}&P(0 \le Z \le z) \\ &= P(Z \le z) - 0.50 \\&= 0.6985 - 0.50 \\&=0.1985 \end{aligned}.

Lookup 0.1985 on a normal distribution table. The corresponding z-score is 0.52. (In other words, P(0 \le Z \le 0.52) = 0.1985.)

Given that

  • z = 0.52,
  • x =5, and
  • \sigma = 2,

Solve the equation \displaystyle z = \frac{x - \mu}{\sigma} for the mean, \mu:

\displaystyle 0.52 = \frac{5 - \mu}{2}.

\mu = 5 - 2 \times 0.52 = 3.96.

3 0
3 years ago
Can someone help me plz!!
ollegr [7]

Answer:

MRT or TRM

Step-by-step explanation:

RT and MR meet at point R. Point R must be included in the final angle as well as the endpoints of RT and MR. The resulting angle is thus MRT, or TRM

4 0
2 years ago
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