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2 years ago

Which points are on the axes 5 units from the origin?

Mathematics

1 answer:

Lera25 [3.4K]2 years ago

5 0

Answer:

b) x,e

Step-by-step explanation:

only look at points on the axies and look for which are only 5 points away

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The compound inequality 8.00 ≤ x < 9.50 represents all values, x, for which college students are paid hourly as teacher assis

Orlov [11]

Answer: x≥ 8.00 and x < 9.50 is the correct option.

Step-by-step explanation:

Since, given expression 8.00 ≤ x < 9.50

Where x is the value for which college students are paid hourly as teacher assistants.

so, x is greater than or equal to 8.00

Therefore, $x\geq 8.00$

And, x is less than 9.50

Therefore, x < 9.50

On combining both expression we get, $x\geq 8.00$ and $x < 9.50$ .

Thus, Third Option is correct.

3 0

3 years ago

Read 2 more answers

How to solve X^4-19x^2+48=0

ZanzabumX [31]

Two numbers that add up to -19 and multiply to 48 are -16 and -3:

$(x^4-19x^2+48)=0\\(x^2-16)(x^2-3)=0\\(x+4)(x-4)(x^2-3)=0$

So, the solutions come from each parentheses: x+4=0, x-4=0, and x^2-3=0.

x+4=0
x = -4

x-4=0
x = 4

x^2-3=0
x^2 = 3
x = +/- √3

So, the solutions are -4, -√3, √3, and 4.

4 0

3 years ago

At a certain high school, 15 students play stringed instruments and

Aleks [24]

Answer:

0 is the probability that a randomly selected student plays both a stringed and a brass instrument.

Step-by-step explanation:

Given that:

Number of students who play stringed instruments, N(A) = 15

Number of students who play brass instruments, N(B) = 20

Number of students who play neither, N( $A \cup B$ )' = 5

To find:

The probability that a randomly selected students plays both = ?

Solution:

Total Number of students = N(A)+N(B)+N( $A \cup B$ )' =15 + 20 + 5 = 40

(As there is no student common in both the instruments we can simply add the three values to find the total number of students)

As per the venn diagram, no student plays both the instruments i.e.

$N(A\cap B) =0$

Formula for probability of an event E can be observed as:

$P(E) = \dfrac{\text{Number of favorable cases}}{\text {Total number of cases}}$

$P(A\cap B) = \dfrac{N(A\cap B)}{N(U)}\\\Rightarrow \dfrac{0}{40}\\\Rightarrow P(A\cap B) = 0$

So, 0 is the probability that a randomly selected student plays both a stringed and a brass instrument.

5 0

2 years ago

Olivia places her pet frog on a line to observe the frog’s motion. The line is divided into sections that measure 1 centimeter e

zvonat [6]

Answer: 0

Explanation: Displacement is how far something or someone is from the starting point, in this case, the frog, so it moves 18F then 6B then 12B, so 0+18-6-12=0.

Hope this helps! :)

4 0

3 years ago

Many airline companies have begun implementing fees for checked bags. Economic theory predicts that passengers will respond to t

valentinak56 [21]

Answer:

Step-by-step explanation:

a. The hypothesis test is one tailed_____ test.

(Because we check whether sample weight is greater than hence one tailed or right tailed)

The test statistic follows a __t___ distribution.(Because only sample std deviation s is known)

The value of the test statistic is___Mean difference/Std error = $\frac{17.1-16}{\frac{6}{\sqrt{67} } } \\=1.58$ __

b. df = 66

Reject H0 if t ≥ 1.668

c. The p-value is_____0.059444

d. Using the critical value approach, the null hypothesis is _accepted____, because __t <1.668___ Using the p-value approach, the null hypothesis is__accepted___, because__p value <0.05 our significance level.___ Therefore, you __may___ conclude that the mean weight of the airline's passengers' carry-on items has increased after the implementation of the checked-bag fee.

4 0

2 years ago