Evaluate the expression

If a pie is divided evenly into thirds, what can you infer about each slice?

Hunter-Best [27]

Answer:

I can infer that each slice is exactly the same size.

Step-by-step explanation:

If the pie is divided evenly into thirds, then each slice is the exact same size.

4 0

3 years ago

What is the LCM of 30 and 35?

Misha Larkins [42]

Answer: 210

Explanation: To find the least common multiple for the integers 30 and 35, we start by making a factor tree for each of our integers.

So 30 is 10 · 3 and 10 is 5 · 2.

35 is 7 · 5.

Notice that our 5's match up as factors of each of our integers.

When finding the least common multiple, we simply multiply all of our factors together but since the 5's match up,

we only multiply by a 5 once.

So our least common multiple or LCM is

5 · the 2 that doesn't match up · the 3 that

doesn't match up · the 7 that doesn't match up.

So we have 5 · 2 · 3 · 7 or 210.

Work is shown below.

4 0

4 years ago

A 50 ft flagpole is mounted to the top of a building. If the angle of elevation from a spot on the street to the top of the pole

monitta

Answer:

the height of the building = 91.67161722 feet.

Step-by-step explanation:

Suppose the height of the building (BC) = X feet.

A 50 ft flagpole (AB) is mounted to the top of a building.

So, height of the top of flag above the ground (AC) = (X+50) feet.

If the angle of elevation from a spot (P) on the street to the top of the pole is 58 degrees and the angle of elevation from the same spot (P) to the bottom of the pole is 46 degrees.

It means ∠APC = 58° and ∠BPC = 46°.

Considering Right triangle ΔACP, cot(∠APC) = PC / AC.

PC = AC*cot(∠APC) = (X+50)*cot(58°)

Considering Right triangle ΔBCP, cot(∠BPC) = PC / BC.

PC = BC*cot(∠BPC) = X*cot(46°)

We have PC = PC.

X*cot(46°) = (X+50)*cot(58°)

0.965688774 * X = 0.624869351 * (X+50)

(0.965688774 - 0.624869351) * X = 0.624869351 * 50

0.340819422 * X = 31.2434676

X = 31.2434676 / 0.340819422 = 91.67161722 feet.

Hence, the height of the building = 91.67161722 feet.

6 0

3 years ago

Each Saturday morning, Janie works 5 hours and earns $35.75. How much does Janie earn for each hour she works?

BabaBlast [244]

Answer/Step-by-step explanation:

If Janie works 5 Hours and earn $35.75..

And It asking for Janie earn for each hour she works..

Therefore, we have to divide to Find how much Janie earn for each hour.

$35.75/5=$7.15

Hence. Janie earn $7.15 each Hour she work.

[RevyBreeze]

4 0

2 years ago

Read 2 more answers

Researchers obtained a sample of n = 10 children who attended daycare before starting school. The children are given a standardi

attashe74 [19]

Answer:

$t=\frac{55.5-50}{\frac{4.249}{\sqrt{10}}}=4.093$

$p_v =2*P(t_{9}>4.093)=0.00084$

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

We can say that at 5% of significance the true mean is significantly different from 50.

Step-by-step explanation:

Data given and notation

53, 57, 61, 49, 52, 56, 58, 62, 51, 56

We can calculate the sample mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s= \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X))^2}{n-1}}$

And we got:

$\bar X=55.5$ represent the sample mean

$s=4.249$ represent the sample standard deviation

$n=10$ sample size

$\mu_o =50/tex] represent the value that we want to test [tex]\alpha=0.05$ represent the significance level for the hypothesis test (ASSUMED).

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is equal or not to 50 :

Null hypothesis: $\mu = 50$

Alternative hypothesis: $\mu \neq 50$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{55.5-50}{\frac{4.249}{\sqrt{10}}}=4.093$

P-value

We need to calculate the degrees of freedom first given by:

$df=n-1=10-1=9$

Since is a two-sided tailed test the p value would given by:

$p_v =2*P(t_{9}>4.093)=0.00084$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis.

We can say that at 5% of significance the true mean is significantly different from 50.

6 0

4 years ago