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Romashka [77]
3 years ago
11

Mutleys journey began on a boat, 2 feet above sea level. He dove down to twelve feet below sea level and then returned to the su

rface of the ocean. Using paint, google drawings/docs, draw Mutleys journey, labeling his 3 locations. Then order the integers from least to greatest. Please help quick
Mathematics
1 answer:
Lubov Fominskaja [6]3 years ago
6 0

Answer:

Mutley would start at 2 feet above sea level or +2, he would then travel 12 feet below sea level (-12) to then return to the surface of the ocean (0).  The integers from least to greatest:  -12, 0, 2.  Visual:

-12(dive level) * * * * * * * * * * * 0(sea level) * 2(boat)

Step-by-step explanation:

Using positive and negative integers, we can determine Mutleys journey above, below and at the surface of the ocean.  The surface of the ocean represents his 'origin' or starting point, which is 0.  The boat is above the surface or +2.  When Mutley dives below the surface, he is at a negative level of the ocean.  Think of it in terms of a number line - negative numbers are to the left of 0 and positive numbers are to the right of 0.  The further we go to the left on the number line, the lower our number.  In this case, -12 would be furthest to the left, then 0, followed by 2.  

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Help asap show work
katovenus [111]

Answer:  see below

<u>Step-by-step explanation:</u>

17) 5n - 3 > -2n + 4     and    9 - 4n ≥ -2n + 1

   <u> +2n   </u>    <u> +2n      </u>              <u>    +2n </u>   <u>+2n      </u>

   7n - 3   >     4                    9 - 2n  ≥       1

        +3         +3                  -9                 -9

    7n       >      7                       -2n   ≥     -8

  <u> ÷7     </u>    <u>     ÷7  </u>                   <u> ÷ -2 </u>  ↓  <u> ÷ -2 </u>

          n  >     1         and            n   ≤   4                  

Graph:   o----------------- ·

              1                     4

18) 9k - 2 ≤ 9 + 10k < 9 - 4k

     9k - 2 ≤ 9 + 10k      and     9 + 10k < 9 - 4k

  <u> -10k     </u>    <u>     -10k </u>               <u>       +4k </u>   <u>    +4k  </u>

     -k - 2 ≤ 9                             9 + 14k < 9

   <u>      +2</u>   <u>+2         </u>                 <u> -9          </u>  <u>-9         </u>

     -k      ≤  11                                  14k  <  0

 <u> ÷ -1    </u>   ↓<u> ÷ -1  </u>                      <u>       ÷14 </u>    <u>÷14      </u>

          k  ≥ -11                and               k < 0

Graph:    · -----------------o

              -11                  0    

19) 4 - n ≤ 10 + 5n ≤ 6n + 1

     4 - n ≤ 10 + 5n       and     10 + 5n ≤  6n + 1

   <u>    -5n </u>   <u>      -5n  </u>                 <u>      -6n </u>   <u>-6n     </u>

   4 - 6n ≤ 10                           10  -  n  ≤       1    

  <u>-4        </u>  <u> -4            </u>              <u> -10        </u>    <u>  -10  </u>

        -6n ≤ 6                                    -n ≤   -9

   <u>  ÷ -6   </u>↓ <u>÷ -6     </u>                    <u>    ÷ -1 </u>↓<u> ÷ -1    </u>

           n ≥ -1              and                 n ≥ 9

Graph:    · -----------------→

              -9                  

20) 4p - 5 ≤ 2p + 9 ≤ 4p + 7

     4p - 5 ≤ 2p + 9     and     2p + 9 ≤ 4p + 7

  <u> -2p       </u>   <u>-2p    </u>                 <u> -4p    </u>   <u>-4p      </u>

    2p - 5 ≤         9                -2p + 9  ≤        7    

    <u>      +5   </u>  <u>    +5    </u>             <u>        -9  </u>    <u>     -9  </u>

       2p  ≤       14                   -2p       ≤      -2

   <u>  ÷ 2   </u>   <u>    ÷ 2     </u>               <u>÷ -2    </u> ↓<u>    ÷ -2    </u>

         p ≤       7          and         p    ≥      1

Graph:    · ----------------- ·

              1                    7

6 0
3 years ago
Suppose the time a child spends waiting at for the bus as a school bus stop is exponentially distributed with mean 7 minutes. De
Gala2k [10]

Answer:

The probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.

Step-by-step explanation:

Let the random variable <em>X</em> represent the time a child spends waiting at for the bus as a school bus stop.

The random variable <em>X</em> is exponentially distributed with mean 7 minutes.

Then the parameter of the distribution is,\lambda=\frac{1}{\mu}=\frac{1}{7}.

The probability density function of <em>X</em> is:

f_{X}(x)=\lambda\cdot e^{-\lambda x};\ x>0,\ \lambda>0

Compute the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning as follows:

P(6\leq X\leq 9)=\int\limits^{9}_{6} {\lambda\cdot e^{-\lambda x}} \, dx

                      =\int\limits^{9}_{6} {\frac{1}{7}\cdot e^{-\frac{1}{7} \cdot x}} \, dx \\\\=\frac{1}{7}\cdot \int\limits^{9}_{6} {e^{-\frac{1}{7} \cdot x}} \, dx \\\\=[-e^{-\frac{1}{7} \cdot x}]^{9}_{6}\\\\=e^{-\frac{1}{7} \cdot 6}-e^{-\frac{1}{7} \cdot 9}\\\\=0.424373-0.276453\\\\=0.14792\\\\\approx 0.148

Thus, the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.

6 0
3 years ago
What is the sum of 3/4 + 5/16
guapka [62]
3/4 + 5/16
=12/16 +5/16
=17/16
=1 1/16
4 0
3 years ago
Read 2 more answers
As an estimation we are told £3 is €4.
vova2212 [387]

Answer:

To solve the above problem we will use the unitary method as follows

As estimated  If £ 3 is equivalent to € 4

Then, £ 1 will be equivalent to = € \frac{4}{3}

£ 64.60 will be equivalent to = € \frac{4}{3} \times 64.60 = 1.3333 \times 64.60 = 86.1311

Now you have to round the answer up to 2 decimal points to get the final answer

€ 86.1311 ≈ € 86.13

Thus, £ 64.60 is approximately equal to  € 86.13.

Step-by-step explanation:

hope this helps if not let me now

3 0
2 years ago
9•(-5)<br><br> A. -45<br> B. -95<br> C. 4<br> D. 45
alexgriva [62]

Answer:

A. -45

Step-by-step explanation:

multiplying a positive an a negative number gives out a negative result

so when we multiple 9 with -5 the result is -45

3 0
2 years ago
Read 2 more answers
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