All categories

3 years ago

Two side lengths of a triangle measure 3 and 9. What is the possible range for the third side length?helppp

Mathematics

1 answer:

faltersainse [42]3 years ago

8 0

Answer:

9-3=6 so range of 6.

Step-by-step explanation:

3,3,3

You might be interested in

There are 820 people at a stadium.65% of them are adults,20% of them are boys,and the rest are girls. How many girls are there a

klemol [59]

100% - 65% - 20% = 15% girls.

15% = 0.15

820 * 0.15 = 123 girls

7 0

4 years ago

Which pair of segments in the figures below are congruent?<br><br>

MrRissso [65]

Answer:

there is actually no pictures

can you requestion it

4 0

3 years ago

Base your answer on Betty is thinking of two consecutive integers whose sum is 41.

mel-nik [20]

Answer:

let the numbers be, x and x+1 ( as it is consecutive number the next number will have 1 more than the number )

sum = 41

so, x + ( x + 1 ) = 41

x + x + 1 = 41

2x + 1 = 41

2x = 41 -1 = 40

2x = 40

x = 40/2 = 20

x = 20

x + 1 = 20 + 1 = 21

so the consecutive numbers whose sum is 40 are, 20 and 21

hope this answer helps you.....

please mark as brainliest... thank you!

6 0

3 years ago

Solve If QR = 6x + 1, PR = 14x - 13, and PQ = 5x - 2, find PR

solong [7]

Answer:

P____Q____R

PR= PQ+ QR

(14x-13) = (5x-2)+(6x+1)

14x-13= 11x – 1

14x – 11x = 13–1

3x = 12

x= 12/ 3

x= 4

PR= 14x – 13 = 14 (4) – 13 = 18 – 13= 5

If you want (PQ , QR ) this is the solution

PQ =5x-2=5(4)-2=20-2=18

QR =6x+1=6(4)+1=24+1=25

I hope I helped you^_^

7 0

3 years ago

A bakery wants to determine how many trays of doughnuts it should prepare each day. Demand is normal with a mean of 5 trays and

Alexeev081 [22]

Answer:

4 trays should he prepared, if the owner wants a service level of at least 95%.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 5

Standard Deviation, σ = 1

We are given that the distribution of demand score is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(X > x) = 0.95

We have to find the value of x such that the probability is 0.95

P(X > x)

$P( X > x) = P( z > \displaystyle\frac{x - 5}{1})=0.95$

$= 1 -P( z \leq \displaystyle\frac{x - 5}{1})=0.95$

$=P( z \leq \displaystyle\frac{x - 5}{1})=0.05$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 5}{1} = -1.645\\x = 3.355 \approx 4$

Hence, 4 trays should he prepared, if the owner wants a service level of at least 95%.

3 0

3 years ago

Two side lengths of a triangle measure 3 and 9. What is the possible range for the third side length?​helppp

Two side lengths of a triangle measure 3 and 9. What is the possible range for the third side length?helppp