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serg [7]
3 years ago
9

Can someone please help me?

Mathematics
1 answer:
Jobisdone [24]3 years ago
8 0

that is a lot of answers though

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4+2(7)= <br><br> 62+8×3= <br><br> 3(2+5)−5(3)+8= <br><br> 24−6+12÷2×3=
Temka [501]

4+2(7)= 18

62+8x3= 86

3(2+5)-5(3)+8= 14

24-6+12÷2x3= 36

Hope this helps you! :D

3 0
3 years ago
What is 5x+4=26 I don’t understand it please help thanks
babunello [35]

Answer: 4.4

Step-by-step explanation:

First you need to take 4 from both sides and then your left with 5x=22

Then you divide from both sides and you get 4.4.

That is your final answer.

3 0
3 years ago
Read 2 more answers
What 3/16 turned into a decimal and 5/48 turned into a decimal
Ira Lisetskai [31]
A way to do it is like this. Numerator divided by denominator so 3 divided by 16 which is .187.

5 divided by 48 is .0146      with the six repeating
7 0
3 years ago
Read 2 more answers
What is the simplified form of ((15xy^2)/(x^2+5x+6))/((5x^2y)/2x^2+7x+3))
skelet666 [1.2K]

Answer:

(27xy +3)/5x^2 +6x

Step-by-step explanation:

((15xy^2 )/ x^2 + 5x + 6 ) × ((2x^2 + 7x + 3) / 5x^2y)

(3y ×(2x+ 7x + 3)) / 5x^2 +6x)

5 0
2 years ago
A survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10. (a)
skad [1K]

Answer:

a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

b) The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

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

Step:-(i)

Given data a survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10

The sample size 'n' = 25

The mean of the sample   x⁻  = $28

The standard deviation of the sample (S) = $10.

Level of significance ∝=0.05

The degrees of freedom γ =n-1 =25-1=24

tabulated value t₀.₀₅ = 2.064

<u>Step 2:-</u>

The 95% of confidence intervals for the average spending

((x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} + t_{\alpha }\frac{S}{\sqrt{n} } )

(28 - 2.064 \frac{10}{\sqrt{25} } ,28 + 2.064\frac{10}{\sqrt{25} } )

( 28 - 4.128 , 28 + 4.128)

(23.872 , 32.128)

a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

b)

Null hypothesis: H₀:μ<30

Alternative Hypothesis: H₁: μ>30

level of significance ∝ = 0.05

The test statistic

t = \frac{x^{-}-mean }{\frac{S}{\sqrt{n} } }

t = \frac{28-30 }{\frac{10}{\sqrt{25} } }

t = |-1|

The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

<u>Conclusion</u>:-

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

8 0
3 years ago
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