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2 years ago

Is -9 3/8 a rational or irrational number

Mathematics

1 answer:

mel-nik [20]2 years ago

3 0

Answer:

Rational

Step-by-step explanation:

-9 ⅜ = -75/8

Every number which can be written in a fraction of integers is rational.

Integers can be negative too

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If a substance decays at a rate of 25% every 10 years, how long will it take 512 grams of the substance to decay to 121.5 grams

Juliette [100K]

Answer:

It will take 50 years to decay from 512 grams to 121.5 grams.

Step-by-step explanation:

The decay formula :

$N=N_0e^{-\lambda t}$

where

N= amount of substance after t time

N₀= initial of substance

t= time.

A substance decays at a rate 25% every 10 years.

So, remaining amount of the substance is = (100%-25%)= 75%

$\frac{N}{N_0}=\frac{75\%}{100\%}=\frac{75}{100}=\frac34$ , t= 10

$N=N_0e^{-\lambda t}$

$\Rightarrow \frac {N}{N_0}=e^{-\lambda t}$

$\Rightarrow \frac34 =e^{-\lambda .10}$

Taking ln both sides

$\Rightarrow ln|\frac34| =ln|e^{-\lambda .10}|$

$\Rightarrow ln|\frac34|=-10\lambda$

$\Rightarrow \lambda=\frac{ ln|\frac34|}{-10}$

Now , N₀= 512 grams, N= 121.5 grams, t=?

$N=N_0e^{-\lambda t}$

$\therefore 121.5=512e^{-\frac{ln|\frac34|}{-10}.t}$

$\Rightarrow 121.5=512e^{\frac{ln|\frac34|}{10}.t}$

$\Rightarrow \frac{121.5}{512}=e^{\frac{ln|\frac34|}{10}.t}$

Taking ln both sides

$\Rightarrow ln|\frac{121.5}{512}|=ln|e^{\frac{ln|\frac34|}{10}.t}|$

$\Rightarrow ln|\frac{121.5}{512}|={\frac{ln|\frac34|}{10}.t}$

$\Rightarrow t=\frac{ln|\frac{121.5}{512}|}{\frac{ln|\frac34|}{10}}$

$\Rightarrow t=\frac{10.ln|\frac{121.5}{512}|}{{ln|\frac34|}}$

⇒t=50 years

It will take 50 years to decay from 512 grams to 121.5 grams.

8 0

3 years ago

Solve for r. - 8x + 14 > 60 OR - 4r + 50 < 58 Choose 1 answer:

hodyreva [135]

R would equal 2 i think

6 0

3 years ago

What facts are true for the graph of the function listed below? Please check all that apply.

lana [24]

Answer:

E, B, A,

Step-by-step explanation:

C is wrong because it isn't decreasing

D is wrong because if X=0 the 5 to the power of zero would be 1 and 2*1= 2 meaning y goes lower than 5

F is wrong because if X=0 the 5 to the power of zero would be 1 and 2*1= 2 meaning the y intercept is 2 not 5

6 0

3 years ago

Read 2 more answers

(−4x 4 +6x 2 +3)−(5x 2 −3)

LekaFEV [45]

Answer:

(-4x4+6x2+3)-(5x2-3)

ok.

-4x4+6x2=2x6

2x6 3

-(+5x2) -(-3)

_____ _____

-3x4 + 6

i think, im not sure but i think its -3x4+6. i might be wrong

3 0

3 years ago

Suppose the expected tensile strength of type-A steel is 103 ksi and the standard deviation of tensile strength is 7 ksi. For ty

ExtremeBDS [4]

Answer:

i So the approximate distribution of $\= X$ is $\mu_{\= X} =103$ and $\sigma_{\= X} = 0.783$

ii So the approximate distribution of $\= Y$ is $\mu_{\= Y} =105$ and $\sigma_{\= Y} = 0.645$

the approximate distribution of $\=X - \= Y$ is $E (\= X - \= Y) = -2$ and $\sigma_{\= X - \=Y}=1.029$

Here we can see that the mean of the approximate distribution is negative which tell us that this negative value of the data for $\=X - \= Y$ sample are more and their frequency occurrence is higher than the positive values

the value of $P(-1 \le \=X - \= Y \le 1)$ is $= -0.1639$

Step-by-step explanation:

From the question we are given that

The expected tensile strength of the type A steel is $\mu_A = 103 ksi$

The standard deviation of type A steel is $\sigma_A = 7ksi$

The expected tensile strength of the type B steel is $\mu_B = 105\ ksi$

The standard deviation of type B steel is $\sigma_B = 5 \ ksi$

Also the assumptions are

Let $\= X$ be the sample average tensile strength of a random sample of 80 type-A specimens

Here $n_a =80$

Let $\= Y$ be the sample average tensile strength of a random sample of 60 type-B specimens.

Here $n_b = 60$

Let the sampling distribution of the mean be

$\mu _ {\= X} = \mu$

$=103$

Let the sampling distribution of the standard deviation be

$\sigma _{\= X} = \frac{\sigma }{\sqrt{n_a} }$

$= \frac{7}{\sqrt{80} }$

$=0.783$

So What this mean is that the approximate distribution of $\= X$ is $\mu_{\= X} =103$ and $\sigma_{\= X} = 0.783$

For $\= Y$

The sampling distribution of the sample mean is

$\mu_{\= Y} = \mu$

$= 105$

The sampling distribution of the standard deviation is

$\sigma _{\= Y} = \frac{\sigma }{\sqrt{n_b} }$

$= \frac{5}{\sqrt{60} }$

$= 0.645$

So What this mean is that the approximate distribution of $\= Y$ is $\mu_{\= Y} =105$ and $\sigma_{\= Y} = 0.645$

Now to obtain the approximate distribution for $\=X - \= Y$

$E (\= X - \= Y) = E (\= X) - E(\= Y)$

$= \mu_{\= X} - \mu_{\= Y}$

$= 103 -105$

$= -2$

The standard deviation of $\=X - \= Y$ is

$\sigma_{\= X - \=Y} = \sqrt{\sigma_{\= X}^2 - \sigma_{\= Y}^2}$

$= \sqrt{(0.783)^2 + (0.645)^2}$

$=1.029$

Now to find the value of $P(-1 \le \=X - \= Y \le 1)$

Let us assume that $F = \= X - \= Y$

$P(-1 \le F \le 1)$ $= P [\frac{-1 -E (F)}{\sigma_F} \le Z \le \frac{1-E(F)}{\sigma_F} ]$

$= P[\frac{-1-(-2)}{1.029} \le Z \le \frac{1-(-2)}{1.029} ]$

$= P[0.972 \le Z \le 2.95]$

$= P(Z \le 0.972) - P(Z \le 2.95)$

Using the z-table to obtain their z-score

$= 0.8345 - 0.9984$

$= -0.1639$

3 0

2 years ago