Answer:
Step-by-step explanation:
<em><u>Given</u></em><u>:</u> A line m is perpendicular to the angle bisector of ∠A. We call this
intersecting point as D. Hence, in figure ∠ADM=∠ADN =90°.
AD is angle bisector of ∠A. Hence, ∠MAD=∠NAD.
<u><em>To Prove</em></u>: <em><u>ΔAMN is an isosceles triangle. i.e any two sides in ΔAMN are</u></em>
<em> </em><em><u>equal. </u></em>
<em><u>Solution</u></em>: Now, In ΔADM and ΔADN
∠MAD=∠NAD ...(1) (∵Given)
AD=AD ...(2) (∵common side)
∠ADM=∠ADN ...(3) (∵Given)
<u><em> Hence, from equation (1),(2),(3) ΔADM ≅ ΔADN</em></u>
( ∵ ASA congruence rule)
⇒<u><em> AM=AN</em></u>
Now, In Δ AMN
AM=AN (∵ Proved)
Hence, ΔAMN is an isosceles triangle.
Answer: 0.11
Step-by-step explanation: (2/3) / 6 = 0.11
Answer:
We need to conduct a hypothesis in order to determine if the mean is greater than specified value, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
For this case the significance is 1%. So we need to find a critical value in the normal standard distribution who accumulates 0.99 of the area in the left and 0.01 in the right and for this case this critical value is:
Step-by-step explanation:
Notation
represent the sample mean
represent the standard deviation for the population
sample size
represent the value that we want to test
represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to determine if the mean is greater than specified value, the system of hypothesis would be:
Null hypothesis:
Alternative hypothesis:
For this case the significance is 1%. So we need to find a critical value in the normal standard distribution who accumulates 0.99 of the area in the left and 0.01 in the right and for this case this critical value is:
Answer:
7 times 5x
Step-by-step explanation:
4 plus x and x has value
Niklas takes a dose of 25 micrograms of a certain supplement each day. The supplement has a half life of 4 hours, meaning that 1/64 of the supplement remains in the body after each day. How much of the supplement is in Niklas's body immediately after the 12th dose? Round your final answer to the nearest hundredth.
Answer:
The amount of the supplement in Niklas body immediately after the 12th dose is 430 micrograms to the nearest hundreth
Step-by-step explanation:
Half life is the time required for an element to decay into half of its initial size.
Given that :
The supplement has a half life of 4 hours, this implies that it decay to half of its size every 4 hours.
∴ there are 6 stages of division in a day.
i.e
The amount of the supplement in Niklas body after the first dose (first day) can be calculated as:
= 
× 25
= 0.390625 micrograms
It is said that he used the supplement daily for 12 days (12th dose),
As such ; we can estimate the amount of the supplement that is in his body immediately after the 12th dose; which is calculated as:
amount in his body per day × number of period for complete decay
= 0.390625 × 11
= 4.296875
≅4.30 micrograms
= 430 micrograms to the nearest hundreth
The amount of the supplement in Niklas body immediately after the 12th dose is 430 micrograms to the nearest hundreth
