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ankoles [38]
3 years ago
12

Dorris has 3 books weighing 2 kilograms each and one container weighing 4 kilograms in a box. How much does the box weigh in kil

ograms?
Mathematics
1 answer:
Sever21 [200]3 years ago
5 0

Answer:

The weight of a box is 10 kg.

Step-by-step explanation:

Weight of each book is 2 kg

So, the weight of 3 books is 6 kg.

Weight of a container is 4 kg

3 books and one container is in a box. We need to find the weight of the box.

It is equal to the sum of weights of 3 books and one container. So,

Total weight = 6 kg + 4 kg

W = 10 kg

So, the weight of a box is 10 kg.

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Please help!!!!!!!!!!! ​
Arisa [49]

Answer:

The ordered pair (15, 12) means that 15 pounds of beans cost $12.

Step-by-step explanation:

Ordered pairs are in the format of (x,y). The first number represents a point's position on the x-axis and the second represents a point's position on the y-axis. Also note that in this graph, the x-axis represents the weight of beans in pounds and the y-axis represents the cost of beans.

Knowing this, (15,12) means that 15 pounds of beans must cost $12.

5 0
3 years ago
Read 2 more answers
0.025 0.25 0.205 0.052 order smallest to biggest
sergij07 [2.7K]
0.025, 0.052, 0.205, 0.25 I believe.  It's been awhile since I've been in Middle School, though. :(
8 0
3 years ago
Solve the inequality 1/4m≤−17 .
Anuta_ua [19.1K]

Answer:

M\leq -68

Step-by-step explanation:

This inequality is quite simple to solve.

Just multiply both sides by four and you have your answer:

4*(1/4m)\leq4*-17

m\leq -68

Hope this helps!

8 0
2 years ago
Shelia's measured glucose level one hour after a sugary drink varies according to the normal distribution with μ = 117 mg/dl and
TiliK225 [7]

Answer:

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

Step-by-step explanation:

To solve this problem, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 117, \sigma = 10.6, n = 6, s = \frac{10.6}{\sqrt{6}} = 4.33

What is the level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L ?

This is the value of X when Z has a pvalue of 1-0.01 = 0.99. So X when Z = 2.33.

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

2.33 = \frac{X - 117}{4.33}

X - 117 = 2.33*4.33

X = 127.1

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

6 0
3 years ago
Order Of operation.
Nikolay [14]

Answer

18(2)+12(2)-4(5)

36+24-20

60-20

40

Step-by-step explanation:

since they had two-18 yard passes and two 12 yard runs, that's the first part of the expression, and then they had four penalties that cost them 5 yards, and that was the last term. When you simplify, you end up with 40 yards gained.

In math

Passes: 18(2)

Runs: 12(2)

Penalties:4(5)

equation

18(2)*12(2)-4(5)

Hope it helped

7 0
3 years ago
Read 2 more answers
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