All categories

3 years ago

Please Help ASAP!!!!

Mathematics

1 answer:

KIM [24]3 years ago

3 0

Answer:

$m\angle M = 98\degree$

Step-by-step explanation:

As we know that:

Sum of the measure of all interior angles of a triangle is always 180°.

$\therefore (3x-16)\degree+(x+6)\degree+ x\degree = 180\degree\\\therefore (5x-10)\degree = 180\degree\\5x-10 = 180\\5x = 180 + 10\\5x = 190\\x= \frac{190}{5}\\x= 38\\\because m\angle M = (3x-16)\degree\\\therefore m\angle M = (3\times 38 -16)\degree\\\therefore m\angle M = (3\times 38 -16)\degree\\\therefore m\angle M = (114 -16)\degree\\\therefore m\angle M = 98\degree$

You might be interested in

Given the box plot, will the mean or the median provide a better description of the center? box plot with min at 11, Q1 at 22. 5

Slav-nsk [51]

The median, because the data distribution is skewed to the right.

Given that

The box plot with the following parameters:

Minimum value at 11,

First Quartile, Q1 at 22.5,

Median at 34.5,

Third Quartile, Q3 at 36,

Maximum value at 37.5.

According to the question

When a distribution has a longer tail to one of its sides, then it is said to be skewed.

Since the tail is longer to the left, it is said to be left-skewed.

The data distribution is skewed to the left because the median (34.5) is closer to the third quartile (36) than to the first quartile (22.5).

For skewed data, we always prefer the median because it is less affected by the outliers and if the mean is chosen, the value of the mean is biased towards the side that has larger values while the median does not get affected by it. So, we choose the median.

For skewed distributions, the median is the best measure of center.

Hence, the median, because the data distribution is skewed to the right.

To know more about Mean click the link given below.

brainly.com/question/2508120

3 0

2 years ago

I need help i also need answers

I am Lyosha [343]

Answer:

Step-by-step explanation:

Given Shane hits 5 HRs out of 100 At-bats,

$\frac{x}{460} = \frac{5}{100} \\\frac{x}{460} (460) = \frac{5}{100} (460)\\x = 23$

5 0

2 years ago

A psychology professor at a university is interested in determining the proportion of undergraduate students at her university w

mario62 [17]

Answer:

No, because the professor’s question had 3 possible answers

6 0

3 years ago

A hair salon completed a survey of 367 customers about satisfaction with service(Dissatisfied, Neutral, Satisfied, or Very Satis

mamaluj [8]

The results of the survey are shown in the table.

It's required to calculate the following probabilities:

a) A customer is neutral

The row labeled as 'Neutral' has a total of 79 customers out of 367 in total.

The required probability is:

$p=\frac{79}{367}$

b) A customer is a walk-in

The first column labeled as 'Walk-in' has a total of 106 customers out of 367. The required probability is

$p=\frac{106}{367}$

c) A customer is dissatisfied and a walk-in.

We have to cross the row Dissatisfied with the column Walk-in. The number of customers that have both categories is 22 out of 367.

The required probability is:

$p=\frac{22}{367}$

d) A customer is very satisfied given he saw a TV ad.

This is a conditional probability, where the total is 111 because is the given condition. The customers that have both characteristics are 33, thus the required probability is:

$p=\frac{33}{111}=\frac{11}{37}$

e) A customer is satisfied or referred.

Total satisfied = 139

Total referred = 150

With both features = 62

Total customers with one or both features: 139 + 150 - 62 = 227

We have to subtract the common feature because it was counted twice.

The required probability is:

$p=\frac{227}{367}$

6 0

1 year ago

The half-life of caffeine in a person’s bloodstream is about 6 hours. If a person’s bloodstream contains 80 milligrams of caffei

Luda [366]

To calculate the remaining caffeine, we use the radioactive decay formula which is expressed as An = Aoe^-kt where An is the amount left after t time, Ao is the initial amount and k is a constant we can calculate from the half-life information. We do as follows:
at half-life,
ln 1/2 = -k(6)
k = 0.12/hr

An = 80e^-0.12(14)
An = 15.87 mg

5 0

3 years ago