All categories

3 years ago

Please help me on this problem.

Mathematics

1 answer:

S_A_V [24]3 years ago

6 0

Answer:

A) 0.6 B) 0.4 C) 0.8

Step-by-step explanation:

all you have to do is count the ammount of sections that fit the description and then divide them by 5. my work for it was:

A) $3/5=0.6$

B) $2/5=0.4$

C) $4/5=0.8$

You might be interested in

Which graph shows a line with the slope of 3/2?

Mnenie [13.5K]

Answer:

Step-by-step explanation:

slope = Rise (y difference) / Run (x difference)

B, Rise = 3 run = 2 slope = 3/2

8 0

2 years ago

NEED HELP ASAP!!!! Okay, I have 5 proportional relationship questions, PLEASE HELP:

Natasha_Volkova [10]

Answer:

The temperature yesterday was -6 because 7-13 is -6.

Step-by-step explanation:

8 0

3 years ago

The expression x4 + 6x3 – 11x2 – 60x + 100 is equivalent to (x – 2)(x – 2)(x + 5)(x + 5). At what points does the graph of the f

nikdorinn [45]

Since y equals the product of the factors, any value for x which makes any factor equal to zero is a point where the graph touches the x-axis.

x=-5 and 2 (technically the points (-5,0) and (2,0))

6 0

3 years ago

Read 2 more answers

Tyler walked 6 miles in 3 hours. What was his walking rate in miles per hour

BARSIC [14]

Answer:

This comes out to 2 mph.

Step-by-step explanation:

Reduce the following fraction:

6 miles

---------------

3 hours

This comes out to 2 mph.

8 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

3 years ago

Please help me on this problem.​

Please help me on this problem.