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bazaltina [42]
2 years ago
10

Write the equation of the line in slope-intercept form.

Mathematics
1 answer:
Leno4ka [110]2 years ago
4 0

Answer:

consider : (7, 0)  \: and  \: (0, 70) \\ m =  \frac{(70 - 0)}{(0 - 7)}  \\ m =  - 10 \\ y = mx + c \\ 0 = ( - 10 \times 7) + c \\ c = 70 \\  \therefore \:  \: y =  - 10x + 70

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Claire is putting up a fence around her rectangular garden. The length is four less than double the width. The perimeter is 22 f
Doss [256]

Answer:

The dimensions are 6 by 5 feet.

Length = 6 feet.

Width = 5 feet.

Step-by-step explanation:

Let the length = L

Let the width = W

Perimeter of a rectangle = 2L + 2W

Translating the word problem into an algebraic equation, we have;

L = 2W - 4 ........equation 1

22 = 2L + 2W .......equation 2

Substituting the value of "L" into equation 2, we have;

22 = 2(2W - 4) + 2W

22 = 4W - 8 + 2W

22 + 8 = 6W

30 = 6W

W = 30/6

Width, W = 5 feet.

To find the length, L

Substituting the value of "W" into equation 1, we have;

L = 2W - 4

L = 2(5) - 4

L = 10-4

Length, L =6 feet

Therefore, the dimensions of the garden are 6 by 5 feet.

5 0
2 years ago
Find the p-value: An independent random sample is selected from an approximately normal population with an unknown standard devi
vladimir1956 [14]

Answer:

(a) <em>p</em>-value = 0.043. Null hypothesis is rejected.

(b) <em>p</em>-value = 0.001. Null hypothesis is rejected.

(c) <em>p</em>-value = 0.444. Null hypothesis is not rejected.

(d) <em>p</em>-value = 0.022. Null hypothesis is rejected.

Step-by-step explanation:

To test for the significance of the population mean from a Normal population with unknown population standard deviation a <em>t</em>-test for single mean is used.

The significance level for the test is <em>α</em> = 0.05.

The decision rule is:

If the <em>p - </em>value is less than the significance level then the null hypothesis will be rejected. And if the <em>p</em>-value is more than the value of <em>α</em> then the null hypothesis will not be rejected.

(a)

The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> > <em>μ₀</em>

The sample size is, <em>n</em> = 11.

The test statistic value is, <em>t</em> = 1.91 ≈ 1.90.

The degrees of freedom is, (<em>n</em> - 1) = 11 - 1 = 10.

Use a <em>t</em>-table t compute the <em>p</em>-value.

For the test statistic value of 1.90 and degrees of freedom 10 the <em>p</em>-value is:

The <em>p</em>-value is:

P (t₁₀ > 1.91) = 0.043.

The <em>p</em>-value = 0.043 < <em>α</em> = 0.05.

The null hypothesis is rejected at 5% level of significance.

(b)

The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> < <em>μ₀</em>

The sample size is, <em>n</em> = 17.

The test statistic value is, <em>t</em> = -3.45 ≈ 3.50.

The degrees of freedom is, (<em>n</em> - 1) = 17 - 1 = 16.

Use a <em>t</em>-table t compute the <em>p</em>-value.

For the test statistic value of -3.50 and degrees of freedom 16 the <em>p</em>-value is:

The <em>p</em>-value is:

P (t₁₆ < -3.50) = P (t₁₆ > 3.50) = 0.001.

The <em>p</em>-value = 0.001 < <em>α</em> = 0.05.

The null hypothesis is rejected at 5% level of significance.

(c)

The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> ≠ <em>μ₀</em>

The sample size is, <em>n</em> = 7.

The test statistic value is, <em>t</em> = 0.83 ≈ 0.82.

The degrees of freedom is, (<em>n</em> - 1) = 7 - 1 = 6.

Use a <em>t</em>-table t compute the <em>p</em>-value.

For the test statistic value of 0.82 and degrees of freedom 6 the <em>p</em>-value is:

The <em>p</em>-value is:

P (t₆ < -0.82) + P (t₆ > 0.82) = 2 P (t₆ > 0.82) = 0.444.

The <em>p</em>-value = 0.444 > <em>α</em> = 0.05.

The null hypothesis is not rejected at 5% level of significance.

(d)

The alternate hypothesis is:

<em>Hₐ</em>: <em>μ</em> > <em>μ₀</em>

The sample size is, <em>n</em> = 28.

The test statistic value is, <em>t</em> = 2.13 ≈ 2.12.

The degrees of freedom is, (<em>n</em> - 1) = 28 - 1 = 27.

Use a <em>t</em>-table t compute the <em>p</em>-value.

For the test statistic value of 0.82 and degrees of freedom 6 the <em>p</em>-value is:

The <em>p</em>-value is:

P (t₂₇ > 2.12) = 0.022.

The <em>p</em>-value = 0.444 > <em>α</em> = 0.05.

The null hypothesis is rejected at 5% level of significance.

5 0
3 years ago
Glenn ate 2 apples a day for a week. In addition to the apples, he ate 3 pears during the week. write the expression that shows
Mama L [17]

Answer:

17 pieces of fruit in total

Step-by-step explanation:

2x7=14+3=17

7 0
2 years ago
The centers for disease control and prevention reported that 25% of baby boys 6-8 months old in the united states weigh more tha
melisa1 [442]

Answer:

0.1971 ( approx )

Step-by-step explanation:

Let X represents the event of weighing more than 20 pounds,

Since, the binomial distribution formula is,

P(x)=^nC_r p^r q^{n-r}

Where, ^nC_r=\frac{n!}{r!(n-r)!}

Given,

The probability of weighing more than 20 pounds, p = 25% = 0.25,

⇒ The probability of not weighing more than 20 pounds, q = 1-p = 0.75

Total number of samples, n = 16,

Hence, the probability that fewer than 3 weigh more than 20 pounds,

P(X

=^{16}C_0 (0.25)^0 (0.75)^{16-0}+^{16}C_1 (0.25)^1 (0.75)^{16-1}+^{16}C_2 (0.25)^2 (0.75)^{16-2}

=(0.75)^{16}+16(0.25)(0.75)^{15}+120(0.25)^2(0.75)^{14}

=0.1971110499

\approx 0.1971

3 0
3 years ago
Please show your work, and have a good day :)
fomenos

Answer:

273x

Step-by-step explanation:

-7x(-8²+25)

8²=64

-7x(-64+25)

-64+25=-39

-7x(-39)

-7×-39=273

273x

7 0
2 years ago
Read 2 more answers
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