Lateral surface area of this gift box

Help me find the missing angles

Keith_Richards [23]

Answer:

b) 105°

d) 130°

Step-by-step explanation:

An external angle has the same measure as the sum of the remote internal angles.

b) ? = 25° +80° = 105°

d) ? = 35° +95° = 130°

_____

This should be fairly obvious if you consider that the adjacent internal angle together with the external angle totals 180°, and the adjacent internal angle together with the other two internal angles totals 180°.

If the two "remote" angles are A and B, and the adjacent internal angle is C, then we have in symbols ...

A + B + C = 180° = ? + C

If we subtract C, then we find ...

A + B = ? . . . . . . the fact we used above

8 0

3 years ago

There are 158 students registered for American History classes. There are twice as many students registered in second period as

Margarita [4]

There are 28 students in first period and 56 students in sceond period and 74 students in third period

<em><u>Solution:</u></em>

Let the number of students in first period be "x"

Let the number of students in second period be "y"

Let the number of students in third period be "z"

<em><u>There are 158 students registered for American History classes.</u></em>

Therefore,

x + y + z = 158 ---------- eqn 1

<em><u>There are twice as many students registered in second period as first period</u></em>

number of students in second period = twice of number of students in first period

y = 2x ------- eqn 2

<em><u>There are 10 less than three times as many students in third period as in first period</u></em>

number of students in third period = 3 times number of students in first period - 10

z = 3x - 10 ------ eqn 3

<em><u>Substitute eqn 2 and eqn 3 in eqn 1</u></em>

x + 2x + 3x - 10 = 158

6x = 168

<em><u>Substitute x = 28 in eqn 2</u></em>

y = 2(28)

<em><u>Substitute x = 28 in eqn 3</u></em>

z = 3(28) - 10

z = 84 - 10

Thus there are 28 students in first period and 56 students in sceond period and 74 students in third period

5 0

2 years ago

Midpoint of the segment between the points (−5,13) and (6,4)

Leokris [45]

Answer: $(0.5,8.5)$

Step-by-step explanation:

We need to use the following formula to find the Midpoint "M":

$M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Given the points (-5,13) and (6,4) can identify that:

$x_1=-5\\x_2=6\\\\y_1=13\\y_2=4$

The final step is to substitute values into the formula.

Therefore, the midpoint of the segment between the points (-5,13) and (6,4) is:

$M=(\frac{-5+6}{2},\frac{13+4}{2})\\\\M=(\frac{1}{2},\frac{17}{2})\\\\M=(0.5,8.5)$

6 0

2 years ago

Read 2 more answers

Write the ordered pair that represents MP. Then find the magnitude of MP.

jarptica [38.1K]

Answer:

(23,-4): $\sqrt{545}$ units

Step-by-step explanation:

We are given that M(-19,4)and P(4,0)

We have to find the ordered pair that represents MP and find the magnitude of MP.

MP=P-M

MP=(4,0)-(-19,4)=(4+19,0-4)

MP=(23,-4)

Magnitude of MP= $\sqrt{(23)^2+(-4)^2}$

Magnitude of MP= $\sqrt{529+16}$

Magnitude of MP= $\sqrt{545}$ units

Hence, .option c is true.

Answer:c.(23,-4): $\sqrt{545}$ units

5 0

3 years ago

A population has a mean of 200 and a standard deviation of 50. Suppose a sample of size 100 is selected and x is used to estimat

zmey [24]

Answer:

a) 0.6426 = 64.26% probability that the sample mean will be within +/- 5 of the population mean.

b) 0.9544 = 95.44% probability that the sample mean will be within +/- 10 of the population mean.

Step-by-step explanation:

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

Normal probability distribution

When the distribution is normal, we use 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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .