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babymother [125]
1 year ago
14

-19>g-24

e=" - 19 \ \textgreater \ g - 24" alt=" - 19 \ \textgreater \ g - 24" align="absmiddle" class="latex-formula">whats the answer
Mathematics
1 answer:
KatRina [158]1 year ago
6 0
\begin{gathered} -19>g-24 \\ -19+24>g \\ 5>g \end{gathered}

start by adding 24 on both sides and simplify

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What is the remainder when (x3 − 7x2 − 18x + 42) is divided by (x + 3)?
Vitek1552 [10]
Look at the picture.
The remainder is 6.

3 0
3 years ago
Read 2 more answers
A researcher wanted to determine whether certain accidents were uniformly distributed over the days of the week. The data show t
Cloud [144]

Answer:

Chi-square = 4.600

Degrees of freedom= 6

there is no evidence to reject hypothesis.

Step-by-step explanation:

H₀ : uniformly distributed

H₁: not uniformly distributed                            

Suppose some value of observed or expected frequency  

Observed frequency : 39 40 30 40 41 49 41     Total: 280

Expected frequency  :  40 40 40 40 40 40 40  Total: 280

Calculated the chi-square value

Chi-square = sigma {(O-E)^2/E}

= 0.025+0.000+2.500+0.000+0.025+2.025+0.025  

 4.600

Degrees of freedom=d f = n-1 = 7-1 = 6

Chi-square for 6 d.f at alpha = 0.05 is 12.59

Chi-square of calculated value < its critical value at alpha = 0.05, we see that there is no evidence to reject hypothesis.

The frequency of Sat is wrongly taken as 41 instead of 51

8 0
3 years ago
Please help!!!!!!!!!!
juin [17]
A: -3.25 or -13/4,b:-13/4?????
8 0
3 years ago
PLS HELP!! Jason bought glass vases for his booth at the fair. If he bought 13 vases and the total weight was 34.97 lb, how much
Stels [109]

Answer:

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6 0
3 years ago
-2=3(y-2) how do you solve this problem for y​
Ksivusya [100]

Answer:

y= 4/3

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

−2=3(y−2)

−2=(3)(y)+(3)(−2)(Distribute)

−2=3y+−6

−2=3y−6

Step 2: Flip the equation.

3y−6=−2

Step 3: Add 6 to both sides.

3y−6+6=−2+6

3y=4

Step 4: Divide both sides by 3.

3y /3 = 4 /3

y= 4/3

Answer:

y= 4/3

8 0
3 years ago
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