Which point is located at (-7, -10)? point N point O point A point E

A number line going from 0 to 4. Numbers 0, 1, 2, 3, and 4 are marked on the line. Marks divide the spaces between each number i

olga nikolaevna [1]

Answer:

8 pieces of string

Step-by-step explanation:

Given that :

Total yards of string = 4

Length in yd of each string = 1/2 yards

Number of pieces of string Ms. Cole has :

Total. Yards / length in yard of each string

4 / (1/2)

4 * (2 /1)

= 8/1

= 8 pieces of string.

6 0

2 years ago

Candice is researching a career as a electrician

kicyunya [14]

More information lol???

8 0

3 years ago

Read 2 more answers

Given the linear system:

azamat

Answer:

y=x+1/2z+-1/2

Step-by-step explanation:

Let's solve for y.

2x−2y+z=1

Step 1: Add -2x to both sides.

2x−2y+z+−2x=1+−2x

−2y+z=−2x+1

Step 2: Add -z to both sides.

−2y+z+−z=−2x+1+−z

−2y=−2x−z+1

Step 3: Divide both sides by -2.

-2y/-2 = -2-z+1/-2

y=x +1/2z+-1/2

3 0

3 years ago

I am a table that uses numbers to record data

Delicious77 [7]

Answer:

Tally table

Step-by-step explanation:

Frequency table- a table that uses numbers to record data. 3. Tally table- a table that uses tally marks to record data bran-list please i answered first i will follow you and answer more of your questions almost expert please

8 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