Select all that apply.

Find the missing side lengths. Leave your answers as radicals in simplest form. Show your work to support your answer

kotegsom [21]

Answer

x=2 and y=2√2

Step-by-step explanation:

$\sin( \alpha ) = \frac{2 \sqrt{2} }{x } \\ x = \frac{2 \sqrt{2} }{ \sin(45) } \\ x = 4 \\ {x}^{2} = {y}^{2} + 2 \sqrt{2} \\ {4}^{2} = {y}^{2} + (2 \sqrt{2 \: )}^{2} \\ 16 = {y}^{2} + 8 \\ 16 - 8 = {y}^{2} \\ y = \sqrt{8} \\ y = 2 \sqrt{2}$

3 0

2 years ago

-10(x+12)-11x.........................,............

NNADVOKAT [17]

-10x -120-11x
combine like terms so -10-11= -21x-120

7 0

3 years ago

Louise measured the perimeter of her rectangular scrapbook to be 156 cm. If the scrapbook is 43 cm wide, how long is the scrapbo

Montano1993 [528]

Answer:

L=35cm

Step-by-step explanation:

Perimeter is 2(L+w)

156cm=2*(43+L)

L=35cm

8 0

3 years ago

3. You roll two fair dice, a white one and a red one. (a) Find P(5 or 6 on the white die and an odd number on the red die) (b) F

Elden [556K]

Answer:

a. P(5 or 6 on the white die and an odd number on the red die) = 1/6

b. P(odd number on the white die and 5 or 6 on the red die) = 1/6

c. P(5 or 6 on the white die and an odd number on the red die) or P(odd number on the white die and 5 or 6 on the red die) = 1/3

Step-by-step explanation:

(a) P(5 or 6 on the white die and an odd number on the red die)

Step 1: P(5 or 6 on the white die) = 1/6 + 1/6 = 1/3

Step 2: P(odd number on the red die) = 1/2

Step 3: P(5 or 6 on the white die and an odd number on the red die) = 1/3 × 1/2 = 1/6

(b) Find P(odd number on the white die and 5 or 6 on the red die).

Step 1: P(odd number on the white die) = 1/2

Step 2: P(5 or 6 on the red die) = 1/6 + 1/6 = 1/3

Step 3: P(odd number on the white die and 5 or 6 on the red die) = 1/2 × 1/3 = 1/6

(c) Find P(5 or 6 on the white die and an odd number on the red die) or P(odd number on the white die and 5 or 6 on the red die).

Step 1: P(5 or 6 on the white die and an odd number on the red die) = 1/6

Step 2: P(odd number on the white die and 5 or 6 on the red die) = 1/6

Step 3: P(5 or 6 on the white die and an odd number on the red die) or P(odd number on the white die and 5 or 6 on the red die) = 1/6 + 1/6 = 1/3

7 0

3 years ago

A manufacturer of chocolate candies uses machines to package candies as they move along a filling line. Although the packages ar

Elenna [48]

Answer:

We conclude that the mean amount packaged is equal to 8.17 ounces.

Step-by-step explanation:

We are given that in a particular sample of 50 packages, the mean amount dispensed is 8.171 ounces, with a sample standard deviation of 0.052 ounces.

Let $\mu$ = population mean amount packaged.

So, Null Hypothesis, $H_0$ : $\mu$ = 8.17 ounces {means that the mean amount packaged is equal to 8.17 ounces}

Alternate Hypothesis, $H_A$ : $\mu\neq$ 8.17 ounces {means that the mean amount packaged is different from 8.17 ounces}

The test statistics that will be used here is One-sample t-test statistics because we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean amount dispensed = 8.171 ounces

s = sample standard deviation = 0.052 ounces

n = sample of packages = 50

So, the test statistics = $\frac{8.171-8.17}{\frac{0.052}{\sqrt{50} } }$ ~ $t_4_9$

= 0.1359

The value of t-test statistics is 0.1359.

Also, the P-value of test-statistics is given by;

the meaning of the p-value is that the p-value is the probability of obtaining a sample mean that is equal to or more extreme than 0.001 ounces away from8.17 if the null hypothesis is true.

P-value = P( $t_4_9$ > 0.136) = More than 40% {from the t-table}

Since the P-value of our test statistics is more than the level of significance of 0.01, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the mean amount packaged is equal to 8.17 ounces.

6 0

3 years ago