"Five less than three times a number is greater than two-thirds of that number."

Rewrite as a simplified fraction.<br> 3.5 = ?

Allushta [10]

Answer: 3 1/2 ( mixed number)

Step-by-step explanation:

First you have to put 3.5 into na fraction

In a fraction it is 35/10
as a mixed number, it would be: 3 5/10 ( a mixed number )
Bur we still have to simplify so simplified would be: 3 1/2 ( mixed number)

7 0

3 years ago

Please help solve this

MArishka [77]

Answer:

The answer to the question provided is 3.

8 0

3 years ago

This symbol > means that one number is _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ or equal to another number.

stellarik [79]

Answer:

gratr

Step-by-step explanation:

4 0

3 years ago

Which expression is equivalent to 4^-5 • 3^-5 ?

valkas [14]

The expression equivalent to 4^-5 • 3^-5 is 12^-5

<h3>What are equivalent expressions?</h3>

Equivalent expressions are simply known as expressions with the same solution but different arrangement.

Given the index expressions;

4^-5 • 3^-5

Using the exponent rule, the two values are have different bases but the same exponent and thus, we multiply the bases and leave the exponents the same way.

This can be written as;

4(3) ^ -5

expand the bracket

12^-5

Thus, the expression equivalent to 4^-5 • 3^-5 is 12^-5

Learn more about equivalent expressions here:

brainly.com/question/24734894

#SPJ1

7 0

2 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

vredina [299]

Answer:

c) P(270≤x≤280)=0.572

d) P(x=280)=0.091

Step-by-step explanation:

The population of bearings have a proportion p=0.90 of satisfactory thickness.

The shipments will be treated as random samples, of size n=500, taken out of the population of bearings.

As the sample size is big, we will model the amount of satisfactory bearings per shipment as a normally distributed variable (if the sample was small, a binomial distirbution would be more precise and appropiate).

The mean of this distribution will be:

$\mu_s=np=500*0.90=450$

The standard deviation will be:

$\sigma_s=\sqrt{np(1-p)}=\sqrt{500*0.90*0.10}=\sqrt{45}=6.7$

We can calculate the probability that a shipment is acceptable (at least 440 bearings meet the specification) calculating the z-score for X=440 and then the probability of this z-score:

$z=(x-\mu_s)/\sigma_s=(440-450)/6.7=-10/6.7=-1.49\\\\P(z>-1.49)=0.932$

Now, we have to create a new sampling distribution for the shipments. The size is n=300 and p=0.932.

The mean of this sampling distribution is:

$\mu=np=300*0.932=279.6$

The standard deviation will be:

$\sigma=\sqrt{np(1-p)}=\sqrt{300*0.932*0.068}=\sqrt{19}=4.36$

c) The probability that between 270 and 280 out of 300 shipments are acceptable can be calculated with the z-score and using the continuity factor, as this is modeled as a continuos variable:

$P(270\leq x\leq280)=P(269.5$

d) The probability that 280 out of 300 shipments are acceptable can be calculated using again the continuity factor correction:

$P(X=280)=P(279.5$

8 0

3 years ago