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iris [78.8K]
2 years ago
7

I'm so confused please help!!!! ( It's not A )

Mathematics
1 answer:
alexgriva [62]2 years ago
4 0

I am pretty sure it is C I hope this helps you.:)

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Find the missing side lengths. Answers are in simplest radical form with the denominator rationalized.
Maksim231197 [3]

Answer:

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8 0
3 years ago
Read 2 more answers
The number of mosquitoes M(x), in millions, in a certain area depends on the June rainfall x, in inches: M(x) = 13 x - x2. What
My name is Ann [436]

Answer:

The number of mosquitoes is maximum for 6.5 inches rainfall.

Step-by-step explanation:

The given function is

M(x)=13x-x^2                   .... (1)

where, M(x) is number of mosquitoes in millions and x the June rainfall in inches.

We need to find the rainfall that produces the maximum number of mosquitoes.

Differential the above function with respect to x.

M'(x)=13-2x               .... (2)

Equate first derivative equal to 0.

13-2x=0

2x=13

x=\frac{13}{2}=6.5

Differential function (2) with respect to x.

M''(x)=-2

Double derivative is negative. So, the value of function is maximum at x=6.5.

Therefore, the number of mosquitoes is maximum for 6.5 inches rainfall.

4 0
3 years ago
Show that x = -2 is a solution of 3x² + 13x + 14 = 0​
Oliga [24]

Answer:

x= -2 is correct answer

Step-by-step explanation:

3(-2×-2)+13(-2)+14=0

3(4)+13(-2)+14=0

12+(-26)+14=0

12-26+14=0

12+14-26=0

26-26=0

0=0

3 0
3 years ago
A cookie factory monitored the number of broken cookies per pack yesterday.
trapecia [35]

Answer:

Confidence Interval - 2.290 < S < 2.965

Step-by-step explanation:

Complete question

A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 50 cookies. The mean is 23.33 and the standard deviation is 2.6. Construct a 80% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies.

Solution

Given  

n=50

x=23.33

s=2.6

Alpha = 1-0.80 = 0.20  

X^2(a/2,n-1) = X^2(0.10, 49) = 63.17

sqrt(63.17) = 7.948

X^2(1 - a/2,n-1) = X^2(0.90, 49) = 37.69

sqrt(37.69) = 6.139

s*sqrt(n-1) = 18.2

s\sqrt{\frac{n-1}{X^2 _{(n-1), \frac{\alpha }{2} } } \leq \sigma \leq s\sqrt{\frac{n-1}{X^2 _{(n-1), 1-\frac{\alpha }{2} } }

confidence interval:

(18.2/7.948) < S < (18.2/6.139)

2.290 < S < 2.965

8 0
2 years ago
hellllpp, with both number 1 and 2, first one to answer gets brainliest, be sure of your answer please!​
Fed [463]

Answer:pretty sure 1 is 23 and 2 is 211200

Step-by-step explanation:

1. 92/4 is 23

2. Just used a calculator on

4 0
2 years ago
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