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bixtya [17]
3 years ago
13

What is the range for the amount of hours studied? 6 18 4 12

Mathematics
1 answer:
Ne4ueva [31]3 years ago
5 0

Answer:

6

Step-by-step explanation:

You take the most amount studied (18) and the least amount studied (12) and subtract them and that would be the range

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John gave away some money. He has $12.00 left. If m represents the money he gave away, which expression represents the money he
makvit [3.9K]

Answer:

the answer would be d

Step-by-step explanation:

you have to divide $12.00 by the amount of money or m

3 0
2 years ago
Some researchers have conjectured that stem-pitting disease in peach-tree seedlings might be controlled with weed and soil treat
stiks02 [169]

Answer:

Part A

b. 14.6 ± 7.38

Part B

b. 3.43

Part C

a. P-value < 0.01

Part D

b. There is sufficient evidence to reject the null hypothesis

Step-by-step explanation:

Part A

The given data are;

The number of seedlings in the field = 20

The number of seedlings selected to receive herbicide A = 10

The number of seedlings selected to receive herbicide B = 10

The height in centimeters of seedlings treated with Herbicide A, \overline x _1 = 94.5 cm

The standard deviation, s₁ = 10 cm

The height in centimeters of seedlings treated with Herbicide B, \overline x _2 = 109.1 cm

The standard deviation, s₂ = 9 cm

The 90% confidence interval for μ₂ - μ₁, is given as follows;

\left (\bar{x}_{2}- \bar{x}_{1}  \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

The critical-t at 95% and n₁ + n₂ - 2 degrees of freedom is given as follows;

The degrees of freedom, df = n₁ + n₂ - 2 = 10 + 10 - 2 = 18

α = 100% - 90% = 10%

∴ For two tailed test, we have, α/2 = 10%/2 = 5% = 0.05

t_{(0.025, \, 18)} = 1.734

C.I. = \left (109.1- 94.5 \right )\pm 1.734 \times \sqrt{\dfrac{10^{2}}{10}+\dfrac{9^{2}}{10}}

C.I. ≈ 14.6 ± 7.37714603353

The 90% C.I. ≈ 14.6 ± 7.38

b. 14.6 ± 7.38

Part B

With the hypotheses are given as follows;

H₀; μ₂ - μ₁ = 0

Hₐ; μ₂ - μ₁ ≠ 0

The two sample t-statistic is given as follows;

t=\dfrac{(\bar{x}_{2}-\bar{x}_{1})}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}+\dfrac{s _{2}^{2}}{n_{2}}}}

t-statistic=\dfrac{(109.1-94.5)}{\sqrt{\dfrac{10^{2} }{10}+\dfrac{9^{2}}{10}}} \approx 3.43173361147

The two sample t-statistic ≈ 3.43

b. 3.43

Part C

From the t-table, the p-value, we have, the p-value < 0.01

a. P-value < 0.01

Part D

Given that a significance level of 0.05 level is used and the p-value of 0.01 is less than the significance level, there is enough statistical evidence to reject the null hypothesis

b. There is sufficient evidence to reject the null hypothesis.

4 0
2 years ago
Write two expressions to show w increased by 4.
iren2701 [21]

Answer:

4+w and w+4

Step-by-step explanation:

I used communative property where you simply flip flop the answers

5 0
3 years ago
Delta furnishings received an invoice dated may 10 for a shipment of goods received june 21. The invoice was for a $8400.00 less
djyliett [7]

Answer:

The right solution is:

(a) $1,940

(b) $2,813

Step-by-step explanation:

Given:

Invoice amount

= $8,400

Discount 1

= 33.33%

Discount 2

= 12.5%

So,

The net payable amount will be:

= 8400\times  [1 - \frac{100}{300} ]\times [1 - \frac{12.5}{100} ]

= 8400\times \frac{2}{3}\times 0.875

= 4900 ($)

Now,

(a)

Amount paid will be:

= 2000\times 0.97

= 1940 ($)

Balance still to be paid will be:

= 4900-2000

= 2900 ($)

(b)

Amount paid will be:

= (4900- 2000)\times 0.97

= 2900\times 0.97

= 2813 ($)

Balance still to be paid will be:

= 2000 ($)

Thus the above solution is the appropriate one.

6 0
2 years ago
Tom is the deli manager at a grocery store. He needs to schedule employees to staff the deli department for no more than 260 per
Rufina [12.5K]

Inequality: 20+40n≤260

Graphing inequality is in the picture I posted.

7 0
3 years ago
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