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d1i1m1o1n [39]
4 years ago
11

455 is divisible by 10. True or false

Mathematics
2 answers:
Yuliya22 [10]4 years ago
7 0

Answer:

true and the answer would be 45.5

Step-by-step explanation:

andrew-mc [135]4 years ago
7 0

Answer:

Depends on the question, if it's asking generally whether it could be divided by 10, then yes. However, if decimals or remainders are not allowed, then no, because the quotient of 455 divided by ten is 45.5.

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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Allegiant Airlines charges a mean base fare of $89. In addition, the airline charges for making a reservation on its website, ch
lara31 [8.8K]

Answer:

a) (<em>μ</em>) = $128

b) 0.9472

c)  0.6671

Step-by-step explanation:

Given that:

Allegiant Airlines charges a mean base fare of $89.

this implies that: mean base fare = $89.

The question proceeds by stating the additional charge on  its website, checking bags, and inflight beverages.

so , additional charges turns out to be = $39 per passenger

Now, Suppose a random sample of 60 passengers is taken

random sample (n) = 60

The population standard deviation of total flight cost is known to be $40

standard deviation (σ) = 40

Question (a) says; we should find the population mean cost per flight

To determine that; we have to consider the total sum(<em>μ</em>) of the mean base fare with the mean additional charges.

Population mean cost per flight (<em>μ</em>)  =  mean base fare + mean additional charges

(<em>μ</em>) = $89 + $39

(<em>μ</em>) = $128

b)

What is the probability the sample mean will be within $10 of the population mean cost per flight (to 4 decimals)?

To determine that; we have:

P(128 - 10 ≤ Х ≤ 128 +10)

= P(118 ≤ Х ≤  138)

= P[\frac{118-128}{40/\sqrt{60} }\leq  \frac{X- \delta }{\alpha /\sqrt{n} }\leq  \frac{138-128}{40\sqrt{60} }]               (where \delta  =  <em>μ  </em>and ∝ = σ )

= P[-1.9365\leq z +1.9365]

= P[z\leq 1.9365]-P[z\leq -1.9365]

Using Excel Command to approach this process, we have;

= 0.9736 - 0.0264

= 0.9472     (to four decimal places)

∴ the probability that the sample mean will be within $10 of the population mean cost per flight = 0.9472

c)

What is the probability the sample mean will be within $5 of the population mean cost per flight (to 4 decimals)?

We wll have to go through the process like the one attempted above in question (b);

So;

P(128 - 5 ≤ Х ≤ 128 + 5)

= P(123 ≤ Х ≤  133)

=  P[\frac{123-128}{40/\sqrt{60} }\leq  \frac{X- \delta }{\alpha /\sqrt{n} }\leq  \frac{133-128}{40\sqrt{60} }]                         (where \delta  =  <em>μ  </em>and ∝ = σ )

= P[-0.9682\leq z +0.9682]

= P[z\leq 0.9682]-P[z\leq-0.9682]

Computing these data in Excel; we have

= 0.8335 -0.1665

= 0.6671         (to 4 decimal places)

∴  the probability that the sample mean will be within $5 of the population mean cost per flight.

5 0
4 years ago
PLEASE HELP
Virty [35]

<u>Answer-</u>

\boxed{\boxed{\dfrac{\widehat{BC}}{\widehat{DE}}=\dfrac{3}{4}}}

<u>Solution-</u>

We know that arc length is the product of radius and central angle in radian.

i.e \text{Arc length}=\text{Radius}\times \text{Central angle}

Here,

\theta_{BC}=1.18\ rad,\ \theta_{DE}=2.36\ rad\\\\AD=\dfrac{2}{3}AB\Rightarrow AB=\dfrac{3}{2}AD

So,

\dfrac{\widehat{BC}}{\widehat{DE}}=\dfrac{AB\times 1.18}{AD\times 2.36}

=\dfrac{\frac{3}{2}AD\times 1.18}{AD\times 2.36}

=\dfrac{3\times 1}{2\times 2}

=\dfrac{3}{4}

4 0
3 years ago
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