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3 years ago

I really need someone to give me all the answers I suck at this

Mathematics

2 answers:

WINSTONCH [101]3 years ago

7 0

5. 3 hours and 52 minutes
6. 5 hours and 11 minutes
7. 4 hours and 51 minutes
8. 1 hour and 58 minutes
9. 4 hours and 14 minutes
10. 3 hours and 47 minutes
11. 1 hour and 11 minutes
12. 3 hours and 24 minutes
13. 3 hours and 11 minutes
14. 1 hour and 36 minutes
15. 4 hours and 56 minutes

zavuch27 [327]3 years ago

5 0

3 hours 52 minutes
3 hours 51 minutes
5 hours 51 minutes
3 hours 38 minutes
6 hours 20 minutes
8 hours 47 minutes

You might be interested in

Step 1: 4y − 2y − 4 = −1 + 13

d1i1m1o1n [39]

You didn't list the options
btw
the next step should be

(add 4 to both sides)
2y = 16
or
2y - 4 + 4 = 12 + 4

8 0

3 years ago

The probability that a randomly selected 2 2-year-old male garter snake garter snake will live to be 3 3 years old is 0.98861 0

Mnenie [13.5K]

Answer:

a. $Probability = 0.97735$

b. $Probability = 0.92294$

c. $P(At\ Least\ One) = 1$

No, it is not unusual if at least 1 lives up to 3.

Step-by-step explanation:

Given

Represent the probability that a 2 year old snake will live to 3 with P(Live);

$P(Live) = 0.98861$

Solving (a): Probability that two selected will live to 3 years.

Both snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

$Probability = P(Live)\ and\ P(Live)$

$Probability = 0.98861 * 0.98861$

$Probability = 0.9773497321$

$Probability = 0.97735$ <em>--- Approximated</em>

Solving (b): Probability that seven selected will live to 3 years.

All 7 snakes have a chance of 0.98861 to live up to 3 years.

So, the required probability is:

$Probability = P(Live)^n$

Where $n = 7$

$Probability = 0.98861^7$

$Probability = 0.92294324145$

$Probability = 0.92294$ <em>--- Approximated</em>

Solving (c): Probability that at least one of seven selected will not live to 3 years.

In probabilities, the following relationship exist:

$P(At\ Least\ One) = 1 - P(None).$

So, first we need to calculate the probability that none of the 7 lived up to 3.

If the probability that one lived up to 3 years is 0.98861, then the probability than one do not live up to 3 years is 1 - 0.98861

This gives:

$P(Not\ Live) = 0.01139$

The probability that none of the 7 lives up to 3 is:

$P(None) = P(Not\ Live)^7$

$P(None) = 0.01139^7$

Substitute this value for P(None) in

$P(At\ Least\ One) = 1 - P(None).$

$P(At\ Least\ One) = 1 - 0.01139^7$

$P(At\ Least\ One) = 0.99999999999997513055642436060443621$

$P(At\ Least\ One) = 1$ ---- Approximated

No, it is not unusual if at least 1 lives up to 3.

This is so because the above results, which is 1 shows that it is very likely for at least one of the seven to live up to 3 years

7 0

3 years ago

What is the equation :y=49-7×

Hunter-Best [27]

Y= 7x +49

hope this helps x

8 0

4 years ago

Solve the equation on the interval [0,2π). cos^4x=cos^4xcscx

Luba_88 [7]

$\bf cos^4(x)=cos^4(x)csc(x)\\\\
-----------------------------\\\\
cos^4(x)=cos^4(x)\cfrac{1}{sin(x)}\implies cos^4(x)=\cfrac{cos^4(x)}{sin(x)}
\\\\\\
cos^4(x)sin(x)=cos^4(x)\implies cos^4(x)sin(x)-cos^4(x)=0
\\\\\\
cos^4(x)[sin(x)-1]=0\to 
\begin{cases}
cos^4(x)=0\to x=cos^{-1}(0)\\
----------\\
sin(x)-1=0\\
sin(x)=1\to x=sin^{-1}(1)
\end{cases}$

$\bf \measuredangle x = cos^{-1}(0)\implies \measuredangle x = 
\begin{cases}
\frac{\pi }{2}\\\\
\frac{3\pi }{2}
\end{cases}
\\\\\\
\measuredangle x=sin^{-1}(1)\implies \measuredangle x=\frac{\pi }{2}$

8 0

4 years ago

Shelly has a cell phone plant that cost $9.99 per month plus $0.05 per minute. Her total bill for the month is $25.59. Write an

Varvara68 [4.7K]

$25.59=$9.99+$0.05m. m=minutes

4 0

3 years ago

Read 2 more answers