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Pepsi [2]
3 years ago
8

Pls help ;-;

Mathematics
2 answers:
anyanavicka [17]3 years ago
6 0

Answer:

C

Step-by-step explanation:

A closed circle means inclusive, and -2 is included in values that would make the equation true. Then you just have to test another number greater than and less than -2 to see which way the arrow should point.

-3(-3) + 1 < 7

9 + 1 < 7

10 < 7 FALSE

-3(0) + 1 < 7

1 < 7 TRUE

All values (including -2) will make this equation true

PIT_PIT [208]3 years ago
4 0

Question:

Which number line shows all the values of x that make the inequality

-2x - 7 ≥ 13 true?

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Which expression is equivalent to 5(2+7)?
frosja888 [35]

Answer:

I can help you but first I need the rest of the answers to find out what's it equivalent to.

4 0
3 years ago
Simplify 4+2 squared
Tanya [424]
4+2^2
=4+(2×2)
=4+(4)
=4+4
=8

or

(4+2)^2
=(4+2) × (4+2)
=16+8+8+4
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hope this helps!! please make my answer brainliest to help me out, thx!!
5 0
3 years ago
Read 2 more answers
HELP PLEASE
Georgia [21]

Answer:

\huge\boxed{\sf No}

Step-by-step explanation:

Evan spent 20 hours doing homework last week

25 hours were spent this week.

<u>Let's see what 125% of 20 equals:</u>

= (125 / 100) * 20

= (125 / 10) * 2

= 125 / 5

= 25 hours

But,

<u>It is written that Evan thought he spent 125 % more than the last week which means:</u>

= (125% of 20) + 20

= 25 + 20

= 45 hours

He should've said that he has given 25% more time than the last week.

(<u>Note that:</u> 25 % of 20 equals 5 so 25 % more than last week will be equal to (25 % + 20) = (5+20) = 25)

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
7 0
3 years ago
How do you find the zeros of an equation​
lubasha [3.4K]

Answer:

Step-by-step explanation:

In general, given the function, f(x), its zeros can be found by setting the function to zero. The values of x that represent the set equation are the zeroes of the function. To find the zeros of a function, find the values of x where f(x) = 0.

5 0
2 years ago
Read 2 more answers
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
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