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aleksandrvk [35]
3 years ago
13

Help me pweaseeeeee):

Mathematics
1 answer:
ella [17]3 years ago
4 0

Answer:

perimeter= 29cm

area= 50.125cm² ( If I didn't make a mistake in the calculation)

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Four pounds of apples cost 3.88. what is the price per pound
Inga [223]

Answer:

0.97 cents

Step-by-step explanation:

4/3.88=0.97

price per pound=0.97 cents

BRAINLIEST APPRECIATED!

HOPE YOU ARE HAVING A GOOD DAY!

6 0
3 years ago
What is 4/11 in decimal form?
Rudiy27
4/11 is 0.363636363636
And "36" just keeps repeating
5 0
4 years ago
How many triangles can be constructed with angles measuring 120°, 40°, and 10°?
balu736 [363]
Any kind of Triangle must have 180 degree of angle in total if it is not, then it is not a Triangle.

In short, Your Answer would be Zero

Hope this helps!
6 0
3 years ago
Read 2 more answers
From the photo below can someone solve question 50 please.
Pie
Question 50.

a. Description

You walk 0.5 miles during 10 minutes, stop and wait for the bus during 4 minutes, then ride the bus for 2 miles during 4 minutes.

b. slopes

the slope of each line represents the average speed of every track.

1) first track

slope = 0.5 miles / 10 minutes = 0.05 miles / minutes.

Given that the slope is constant, you walked at a constant speed of 0.05 miles/minute.

2) second track

slope = 0 => you didn,t move (you were waiting the bus)

3) third track

slope = (2.5 - 0.5) miles / (18 min - 14 min) = 2 miles / 4 min = 0.5 miles / min

means the speed of the bus was constant and equal to 0.5 miles / min.
6 0
3 years ago
The distribution of lifetimes of a particular brand of car tires has a mean of 51,200 miles and a standard deviation of 8,200 mi
Orlov [11]

Answer:

a) 0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

b) 0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

c) 0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

d) 0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean

Step-by-step explanation:

Problems of normally distributed distributions are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

\mu = 51200, \sigma = 8200

Probabilities:

A) Between 55,000 and 65,000 miles

This is the pvalue of Z when X = 65000 subtracted by the pvalue of Z when X = 55000. So

X = 65000

Z = \frac{X - \mu}{\sigma}

Z = \frac{65000 - 51200}{8200}

Z = 1.68

Z = 1.68 has a pvalue of 0.954

X = 55000

Z = \frac{X - \mu}{\sigma}

Z = \frac{55000 - 51200}{8200}

Z = 0.46

Z = 0.46 has a pvalue of 0.677

0.954 - 0.677 = 0.277

0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.

B) Less than 48,000 miles

This is the pvalue of Z when X = 48000. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{48000 - 51200}{8200}

Z = -0.39

Z = -0.39 has a pvalue of 0.348

0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.

C) At least 41,000 miles

This is 1 subtracted by the pvalue of Z when X = 41,000. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{41000 - 51200}{8200}

Z = -1.24

Z = -1.24 has a pvalue of 0.108

1 - 0.108 = 0.892

0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.

D) A lifetime that is within 10,000 miles of the mean

This is the pvalue of Z when X = 51200 + 10000 = 61200 subtracted by the pvalue of Z when X = 51200 - 10000 = 412000. So

X = 61200

Z = \frac{X - \mu}{\sigma}

Z = \frac{61200 - 51200}{8200}

Z = 1.22

Z = 1.22 has a pvalue of 0.889

X = 41200

Z = \frac{X - \mu}{\sigma}

Z = \frac{41200 - 51200}{8200}

Z = -1.22

Z = -1.22 has a pvalue of 0.111

0.889 - 0.111 = 0.778

0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean

4 0
3 years ago
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