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3 years ago

Write 5y<3 without exponents

Mathematics

1 answer:

kondor19780726 [428]3 years ago

4 0

Step-by-step explanation:

y<3/5

3/5 is equivalent to 0.6

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Write the polynomial in standard form and identify the x-intercept, as well as the zeros for

likoan [24]

Well, ok, the x intercepts aer where the function equals 0
also
if the x intercepts are r1 and r2 then the factored form of the function is
f(x)=a(x-r1)(x-r2) where a is a constant
well, we have it already in that form

f(x)=(x-3i)(x-3i)
the x intercept is at x=3i
standard form requires expansion
we expand to get
f(x)=x²-6i-9

7 0

3 years ago

I need these answers, please helppp

aliina [53]

Answer:

Prob of d>4: 0.333

Step-by-step explanation:

180*.333=60

4 0

3 years ago

HELP PLSSSSSSSZzzzzzzzzzzzz

Dvinal [7]

Answer:

w equal to 9z

or z equals to w/9

6 0

3 years ago

The point A(5,11) is a reflected across the x axis what are the coordinates of A'?

LiRa [457]

Answer:

A'(5,-11)

Step-by-step explanation:

Your only changing the y if you are reflecting over the x-axis but change the x if you are reflecting over the y-axis.

8 0

3 years ago

Let X be the waiting time for a car to pass by on a country road, where X has an average value of 35 minutes. If the random vari

Gelneren [198K]

Answer:

Probability that the wait time is greater than 37 minutes is 0.3474.

Step-by-step explanation:

We are given that the random variable X is known to be exponentially distributed and X be the waiting time for a car to pass by on a country road, where X has an average value of 35 minutes.

<u><em>Let X = waiting time for a car to pass by on a country road</em></u>

The probability distribution function of exponential distribution is given by;

$f(x) = \lambda e^{-\lambda x} , x >0$ where, $\lambda$ = parameter of distribution.

Now, the mean of exponential distribution is = $\frac{1}{\lambda}$ which is given to us as 35 minutes that means $\lambda = \frac{1}{35}$ .

So, X ~ Exp( $\lambda = \frac{1}{35}$ )

Also, we know that Cumulative distribution function (CDF) of Exponential distribution is given as;

$F(x) = P(X \leq x) = 1 - e^{-\lambda x}$ , x > 0

Now, Probability that the wait time is greater than 37 minutes is given by = P(X > 37 min) = 1 - P(X $\leq$ 37 min)

P(X $\leq$ 37 min) = $1 - e^{-\frac{1}{35} \times 37}$ {Using CDF}

= 1 - 0.3474 = 0.6525

So, P(X > 37 min) = 1 - 0.6525 = 0.3474

Therefore, probability that the wait time is greater than 37 minutes is 0.3474.

4 0

3 years ago