Which expressions are a factor of -48xyz -24xy +40

you have diving lessons every fifth day and swimming lessons every third day today you have both lessons in how many days will y

olganol [36]

Mondays, Tuesdays, Wednesdays.

8 0

3 years ago

The area of the conference table in Mr. Nathan’s office must be no more than 175 ft2. If the length of the table is 18 ft more t

stich3 [128]

The answer i think is. If the length of the table is 18 ft more than the width, x, which - 1447102. ... Thearea of the conference table in Mr. Nathan's office must be no more than 175 ft2 ... Therefore, theinterval 0 < x ≤ 7 represents the possible widths

7 0

3 years ago

A shoemaker makes 15,200$ and 912$ in commission (included in 15,200). What is the percent he made in commission to he whole amo

kodGreya [7K]

The total amount of money he only made is calculated by the formula:

amount made = 15,200 – 912

amount made = 14,288

Therefore the percentage commission was:

% commission = 912 / 14,288 * 100%

% commission = 6.38 %

6 0

3 years ago

Solve using the quadratic formula. Write answers as an exact value and as a decimal rounded to the nearest hundredth.

vichka [17]

Answer:

$\frac{-2±\sqrt{3} }{2}$

dec form: -0.13

Step-by-step explanation:

4 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