All categories

3 years ago

1.If mZA = 4x+12 and x = 15, is ZA acute, obtuse, or right? Explain or justify your answer.

Mathematics

1 answer:

spin [16.1K]3 years ago

6 0

9514 1404 393

Answer:

acute
17

Step-by-step explanation:

1. Put the value of x in the expression and see what the angle measure is.

m∠A = 4x +12 = 4(15) +12 = 60 +12 = 72

The angle is less than 90°, so is acute.

2. The measure of a right angle is 90°, so you have ...

4x +22 = 90

4x = 68 . . . . . subtract 22; next, divide by 4

x = 17

You might be interested in

Given that BCD and BCA form a straight angle, if BAC measures 40 degrees and BCD measures 75 degrees, then the measure of ABC is

Irina-Kira [14]

Answer:

∠ABC= 35°

Step-by-step explanation:

im not sure if u need explanation but comment if u do

6 0

3 years ago

The population of a pigeons in a city is 100 and is growing exponentially at 20% per year. Write a function to represent the pop

NikAS [45]

Answer:

p(t) = 100×1.0153^(12t)
1.53% per month

Step-by-step explanation:

In general, the function will be written ...

p(t) = (initial value)×(growth factor)^t

where t is in units comparable to those applicable to the growth factor. The growth factor is found from ...

growth factor = 1 + growth rate

Here, the growth rate is given as 20% per year, so the growth factor per year is ...

1 +20% = 1.20

The initial value is given as 100, so we can write the exponential function as ...

p(t) = 100×1.20^t

The time period units for t are supposed to be years, but we want to find the growth rate for a month. We can do that by recognizing there are 12 months in a year. In the above equation, we can use (1/12)(12t) in place of t, then figure the growth factor (and growth rate) per month.

p(t) = 100×(1.20^(1/12))^( 12t)

p(t) = 100×1.0153^(12t) . . . . population exponential function

This shows the monthly growth factor is 1.0153, so the monthly rate of change (growth rate) is ...

1.0153 -1 = 0.0153 = 1.53% . . . . monthly rate of change

5 0

3 years ago

Read 2 more answers

Math workshops and final exams: The college tutoring center staff are considering whether the center should increase the number

ludmilkaskok [199]

Answer:

2. Number of workshops attended.

Step-by-step explanation:

The variable of interest for predicting the final exam score and doing regression analysis is workshop attendance. Therefore, the response variable should be the number of workshops attended by each student.

This also agrees with what the college tutoring center staff are considering, which forms the research question: "should the center increase the number of math workshops they offer to help students improve their performance in math classes?"

3 0

4 years ago

The mass of a grain of sand is about 10^-3 gram. About how many grains of sand are in the bag of sand?

nika2105 [10]

there are 10 million grains in that bag of sand this is because one grain equals to 10 power -3 what about 10 kg you change it into grams then you cross multiply and get the answer

8 0

3 years ago

A population has mean 187 and standard deviation 32. If a random sample of 64 observations is selected at random from this popul

Zina [86]

Answer:

11.51% probability that the sample average will be less than 182

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples 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.

Central Limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$