In a load of 3/4 ton of steel rods, 1/8 of them are bent. How many tons of steel rods are bent. Mm

Mr. Smith had 34 of a peach pie left over. He split the leftover pie among his 3 co-workers.

viktelen [127]

Answer:

11.333 infinity

Step-by-step explanation:

when you divide 34 by 3 you then get 11.3 infinity which would also be a repeating decimal

8 0

3 years ago

Graph the solution to this system of inequalities in the coordinate plane.

Vsevolod [243]

By using Geogebra online graphing calculator, the solution to this system of inequalities are plotted in the image attached below.

<h3>What is an inequality?</h3>

An inequality is a mathematical relation that compares two (2) or more integers and variables in an equation based on any of the following:

Less than (<).

Greater than (>).

Less than or equal to (≤).

Greater than or equal to (≥).

Given the following system of inequalities:

3y > 2x + 12 ⇒ 3y - 2x > 12

2x + y ≤ -5

By using Geogebra online graphing calculator, the solution to this system of inequalities are plotted in the image attached below.

Read more on inequalities here: brainly.com/question/24372553

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3 0

1 year ago

For one English test, Benny had to answer 15 questions. Of these questionsBenny answered 60% of them correctly. How many questio

LenaWriter [7]

Answer:

9 correct

Step-by-step explanation: 60% of 15 = 9

6 0

2 years ago

How can you use rates of change to solve a real-life problem like the excessive emission of CO2 in the air? Answer using correct

Norma-Jean [14]

Answer:

rate of change can be used in a real life problem such as the rate that a pool fills up with water.

8 0

2 years ago

An instructor who taught two sections of engineering statistics last term, the first with 25 students and the second with 35, de

Amiraneli [1.4K]

Answer:

a) P=0.1721

b) P=0.3528

c) P=0.3981

Step-by-step explanation:

This sampling can be modeled by a binominal distribution where p is the probability of a project to belong to the first section and q the probability of belonging to the second section.

a) In this case we have a sample size of n=15.

The value of p is p=25/(25+35)=0.4167 and q=1-0.4167=0.5833.

The probability of having exactly 10 projects for the second section is equal to having exactly 5 projects of the first section.

This probability can be calculated as:

$P=\frac{n!}{(n-k)!k!}p^kq^{n-k}= \frac{15!}{(10)!5!}\cdot 0.4167^5\cdot0.5833^{10}=0.1721$

b) To have at least 10 projects from the 2nd section, means we have at most 5 projects for the first section. In this case, we have to calculate the probability for k=0 (every project belongs to the 2nd section), k=1, k=2, k=3, k=4 and k=5.

We apply the same formula but as a sum:

$P(k\leq5)=\sum_{k=0}^{5}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have:

$P(k=0)=0.0003\\P(k=1)=0.0033\\P(k=2)=0.0165\\P(k=3)=0.0511\\P(k=4)=0.1095\\P(k=5)=0.1721\\\\P(k\leq5)=0.0003+0.0033+0.0165+0.0511+0.1095+0.1721=0.3528$

c) In this case, we have the sum of the probability that k is equal or less than 5, and the probability tha k is 10 or more (10 or more projects belonging to the 1st section).

The first (k less or equal to 5) is already calculated.

We have to calculate for k equal to 10 or more.

$P(k\geq10)=\sum_{k=10}^{15}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have

$P(k=10)=0.0320\\P(k=11)=0.0104\\P(k=12)=0.0025\\P(k=13)=0.0004\\P(k=14)=0.0000\\P(k=15)=0.0000\\\\P(k\geq10)=0.032+0.0104+0.0025+0.0004+0+0=0.0453$

The sum of the probabilities is

$P(k\leq5)+P(k\geq10)=0.3528+0.0453=0.3981$

8 0

3 years ago