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3 years ago

Which of the following values are solutions to the inequality 1 + 5x < 4?

Mathematics

1 answer:

Romashka-Z-Leto [24]3 years ago

7 0

Step-by-step explanation:

you cant add 1 to 5x, so subtract 1 on both sides leaving you with 1 + 5x < 4 that equals 5x < 3 -1 -1

then divide 5 from both sides 5x < 3 This leaves 5 5

you with x < 3/5

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At the movie theater child admission is $5.20 and adult admission is $9.90.. On Friday 163 tickets were sold for a total sales o

Jet001 [13]

The number of tickets sold on Friday were 66 children tickets and 97 adult tickets.

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

An equation is an expression that shows the relationship between two or more numbers and variables.

Let x represent the number of children and y represent the number of adult, hence:

5.2x + 9.9y = 1303.5 (1)

Also:

x + y = 163 (2)

From both equations:

x = 66, y = 97

The number of tickets sold on Friday were 66 children tickets and 97 adult tickets.

Find out more on equation at: brainly.com/question/2972832

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1 year ago

Find the solution to the system.<br> y = X-6<br> y = -3x + 2

zmey [24]

Answer:

x=6

Step-by-step explanation:

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2 years ago

Read 2 more answers

Item 16<br> Solve the equation.<br><br> −49f=−3

weqwewe [10]

I think the answer is either 3/49 or about 0.061
hope this helps.

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3 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.

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