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klasskru [66]
3 years ago
12

6th grade math help me pleaseeee...

Mathematics
1 answer:
iren2701 [21]3 years ago
3 0

Answer:

-3.8

Step-by-step explanation:

u = 3.8

u × -1 = 3.8 × -1

-u = -3.8

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How many 3/4 cup servings are in 12 cups of cherries?
AveGali [126]

Answer:

16 serving.

Step-by-step explanation:

Given: 3/4 cup servings.

Now, computing the number of serving in 12 cups of cherries.

Using unitary method to solve.

∴ Number of serving= total\ cups \times number\ of\ serving\ per\ cup

Number of serving in one cup is \frac{4}{3}

Number of serving in 12 cups = 12\times \frac{4}{3} = 16 \ serving

Number of serving in 12 cups= 16\ serving

∴ There are 16 serving is possible in 12 cups of cherries.

8 0
3 years ago
Who is correct Emmitt or Emily
lianna [129]

Answer:

I think Emily is

Step-by-step explanation:

8 0
2 years ago
PLEASE HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP
Anvisha [2.4K]

Answer:

4/18

Step-by-step explanation:

3 0
2 years ago
A random sample of 23 items is drawn from a population whose standard deviation is unknown. The sample mean isx⎯ ⎯ x¯ = 840 and
gayaneshka [121]

Answer:

(832.156, \ 847.844)

Step-by-step explanation:

Given data :

Sample standard deviation, s = 15

Sample mean, \overline x = 840

n = 23

a). 98% confidence interval

$\overline x \pm t_{(n-1, \alpha /2)}. \frac{s}{\sqrt{n}}$

$E= t_{( n-1, \alpha/2 )} \frac{s}{\sqrt n}}

$t_{(n-1 , \alpha/2)} \frac{s}{\sqrt n}$

$t_{(n-1, a\pha/2)}=t_{(22,0.01)} = 2.508$

∴ $E = 2.508 \times \frac{15}{\sqrt{23}}$

  $E = 7.844$

So, 98% CI is

$(\overline x - E, \overline x + E)$

(840-7.844 ,  \ 840+7.844)

(832.156, \ 847.844)

4 0
3 years ago
Suppose that each observation in a random sample of 100 fatal bicycle accidents in 2015 was classified according to the day of t
liraira [26]

Answer:

The calculated χ² =  0.57   does not fall in the critical region χ² ≥  12.59  so we fail to reject the null hypothesis and conclude the proportion of fatal bicycle accidents in 2015 was the same for all days of the week.

Step-by-step explanation:

1) We set up our null and alternative hypothesis as

H0:  proportion of fatal bicycle accidents in 2015 was the same for all days of the week

against the claim

Ha:  proportion of fatal bicycle accidents in 2015 was not the same for all days of the week

2) the significance level alpha is set at 0.05

3) the test statistic under H0 is

χ²= ∑ (ni - npi)²/ npi

which has an approximate chi square distribution with ( n-1)=7-1=  6 d.f

4) The critical region is χ² ≥ χ² (0.05)6 = 12.59

5) Calculations:

χ²= ∑ (16- 14.28)²/14.28 + (12- 14.28)²/14.28 + (12- 14.28)²/14.28 + (13- 14.28)²/14.28 + (14- 14.28)²/14.28 + (15- 14.28)²/14.28 + (18- 14.28)²/14.28

χ²= 1/14.28 [ 2.938+ 5.1984 +5.1984+1.6384+0.0784 +1.6384+13.84]

χ²= 1/14.28[8.1364]

χ²= 0.569= 0.57

6) Conclusion:

The calculated χ² =  0.57   does not fall in the critical region χ² ≥  12.59  so we fail to reject the null hypothesis and conclude the proportion of fatal bicycle accidents in 2015 was the same for all days of the week.

b.<u> It is r</u>easonable to conclude that the proportion of fatal bicycle accidents in 2015 was the same for all days of the week

6 0
3 years ago
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