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DerKrebs [107]
3 years ago
12

ILL MARK BRAINILEST IF YOU ANSWER!!!

Mathematics
1 answer:
ella [17]3 years ago
8 0

Answer:

h_{max}=5.5

Step-by-step explanation:

Given

h = -0.1(d + 1)(d - 12)

Solving (a): The graph

To do this, we plot h on the vertical axis and d on the horizontal

<em>See attachment for graph</em>

Solving (b): The maximum of h

From the attached graph, h has the maximum value when:

d = 4.225

So, the maximum of h is:

h_{max}=5.5

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If observed frequencies are 5,10,15 and expected frequencies are each equal to 10 then chi square value is
lina2011 [118]

Answer:

5

Step-by-step explanation:

Given :

Observed values = 5, 10, 15

Expected freqencies = 10

χ² = Σ(observed - Expected)² / Expected

χ² = (5-10)²/10 + (10-10)²/10 + (15-10)²/10

χ² = 25/10 + 0/10 + 25/10

χ² = 2.5 + 0 + 2.5

χ² = 5

5 0
3 years ago
Gcf of 18m x 18m and 7m
IrinaVladis [17]
The Gcf of 18m x 18m and 7m is 1
6 0
3 years ago
11/44 as a equivalent fraction
Novay_Z [31]

Answer:

1/4

Step-by-step explanation:

11/44 = 1/4

Divide the numerator and the denominator by 11

11 ÷ 11 / 44 ÷ 11

1/4

Hope this helped!

Have a supercalifragilisticexpialidocious day!

8 0
2 years ago
Read 2 more answers
_÷7 = 3 × 1 I dont know what else to but since the question is too short to post
Afina-wow [57]
_ divided by 7 = 3 * 1.    Just do 7*3=21.     so 21 divided by 7 = 3 * 1
7 0
3 years ago
Read 2 more answers
The rate at which radioactive waste is entering the atmosphere at time t is decreasing and is given by Pe^-kt, where P is the in
Maslowich

Answer:

8333.33

Step-by-step explanation:

Given that the  rate at which radioactive waste is entering the atmosphere at time t is decreasing and is given by Pe^-kt, where P is the initial rate. Use the improper integral

∫^[infinity]_0 Pe^-kt dt

with P = 50 to find the total amount of the waste that will enter the atmosphere for each value of k.

k=0.06

Hence the total amount of waste that can  enter the atmosphere when  k =0.06 is given by

\int\limits^{\infty} _0 {50e^{-0.06t} } \, dt\\=-\frac{50}{0.006} e^{-0.06t}

Substitute limits

=8333.33

7 0
3 years ago
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