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Gre4nikov [31]
3 years ago
12

There is an 85% chance that a flight on Bandway Airline will arrive on time. Out of the total of 600 fights, how many are expect

ed to arrive on time ?
Mathematics
2 answers:
Ierofanga [76]3 years ago
5 0
600 * 0.85 = 510 (for those wanting an explanation)
olga55 [171]3 years ago
4 0

Answer: 510 flights will arrive on time

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Which of the following is the equation in slope-intercept form for the line that passes through points (- 4, 1) and \ -3,0\; y =
Akimi4 [234]

Answer:

there y= -x - 3 is correct

Step-by-step explanation:

(-4,1) ; (-3,0)

y= -x - 3

with (-4,1)

1 = -(-4) - 3 = 4 - 3 = 1. left = right

with (-3,0)

0 = -(-3) - 3 = 3 - 3 = 0. left = right

so the line y = -x - 3 is passes through points (-4,1) and (-3,0)

3 0
2 years ago
If tan 3x =<img src="https://tex.z-dn.net/?f=%5Csqrt%7Bx%7D" id="TexFormula1" title="\sqrt{x}" alt="\sqrt{x}" align="absmiddle"
aliya0001 [1]

Answer:

tan3x=√3

tan3x=tan60,tan(180+60)

3x=60,240

:.x=20,80is your answer.

c=

6 0
3 years ago
What is 25% of 40...?
11Alexandr11 [23.1K]

Answer:

10

Step-by-step explanation:

Searched it up

4 0
3 years ago
Describe what transformations of the quadratic parent function of g(x)=2(x-3)^2
Elodia [21]

Answer: There is a horizontal shift of 3 units to the right and a vertical stretch by a factor of 2.

Step-by-step explanation: The (x-3) shifts the graph three units right, think of as for (x)=0 x has to be 0, and for (x-3)=0 x has to be 3, which is 0+3. The vertical stretch multiplies everything by 2, so it should make sense that it's a vertical stretch by a factor of 2.

6 0
2 years ago
Read 2 more answers
A data set about speed dating includes​ "like" ratings of male dates made by the female dates. The summary statistics are nequal
UNO [17]
<h2>Answer with explanation:</h2>

Given : Sample size n= 189

Sample mean : \overline{x}=7.58

Sample standard deviation : s=1.93

Let \mu be the population mean of "like" ratings of male dates made by the female dates.

As per question ,

Null hypothesis : \mu\geq8.00

Alternative hypothesis : \mu , It means the test is a one-tailed t-test. ( we use t-test when population standard deviation is unknown.)

Test statistic:

t_{stat}=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}\\\\=\dfrac{7.58-8.00}{\dfrac{1.93}{\sqrt{189}}}=-2.99

For 0.05 significance and df =188 (df=n-1) p-value = .001582. [By t-table]

Since  .001582< 0.05

Decision: p-value < significance level , that means there is statistical significance, so we reject the null hypothesis.

Conclusion : We support the claim at 5% significance that t the population mean of such ratings is less than 8.00.

4 0
4 years ago
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