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2 years ago

Warm-Up

Mathematics

1 answer:

Sveta_85 [38]2 years ago

4 0

Answer:

Graph # 3

Step-by-step explanation:

-2x + 5y > 15 Let x = 0 solve for y

-2(0) + 5y = 15 change the > to an =

5y = 15

y =3 Point (0, 5) is on the graph

Graph # 3 is correct because the y-intercept is 5

x y

0 3 -2(0) + 5y = 15; 5y = 15

5 5 -2(5) + 5y = 15; -10 + 5y = 15; 5y = 25; y = 5

10 7 -2(10) + 5y = 15; -20 + 5y = 15; 5y = 35; y = 7

The graph > the line is dotted and you will shade above the line

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Natalia bought 8 glitter wands for 50$

kirza4 [7]

Answer:

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3 years ago

Ralph Phillips obtained a 12-month, $1500.00 loan at 12%. His monthly payment is $133.20. For the first payment find:

Paul [167]

Answer:

a. The interest is US$ 8.20

b. The payment to principal is US$ 125.

c. The new balance is: US$ 1,465.20 (including principal and interest) or US$ 1,375 (just the principal).

Step-by-step explanation:

1. Let's review the information given to answer this question correctly:

Loan obtained by Ralph Phillips = US$ 1,500

Time of paying back the loan = 12 months

Interest rate of the loan = 12%

Monthly payment = US$ 133.20

2. Let's find a, b and c.

The principal is US$ 1,500. Dividing it by 12 months, the monthly payment of principal is US$ 125. Therefore,

a. The interest is (133.20 - 125) = US$ 8.20

b. The payment to principal is US$ 125.

c. The new balance is: US$ 1,465.20 (including principal and interest) or US$ 1,375 (just the principal).

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3 years ago

How to do this geometry’s

horsena [70]

Answer:

150 in^3

Step-by-step explanation:

A= B*h / 2

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3 years ago

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Ksju [112]

Answer:

Step-by-step explanation:

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2 years ago

Read 2 more answers

The thickness of metal wires used in the manufacture of silicon wafers is assumed to be normallydistributed with meanμ. To monit

yanalaym [24]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu$ = 10

(b) Alternate Hypothesis, $H_A$ : $\mu\neq$ 10

(c) One-sample t test statistics distribution : T.S. = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

(d) The value of the test statistic is 1.054.

(e) The p-value = 0.1493

(f) We conclude that the mean equals the target value of 10 which means that the output can be considered acceptable as it doesn't differs from the target value of 10.

Step-by-step explanation:

We are given that the thickness of metal wires used in the manufacture of silicon wafers is assumed to be normally distributed with mean μ. To monitor the production process, the thickness of 40 wires is taken.

The output is considered unacceptable if the mean differs from the target value of 10. The 40 measurements yield a sample mean of 10.2 and sample standard deviation of 1.2.

Let $\mu$ = mean thickness of metal wires used in the manufacture of silicon wafers.

(a) Null Hypothesis, $H_0$ : $\mu$ = 10 {means that the mean equals the target value of 10}

(b) Alternate Hypothesis, $H_A$ : $\mu\neq$ 10 {means that the mean differs from the target value of 10}

(c) The test statistics that will be used here is One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\mu$ = sample mean = 10.2

s = sample standard deviation = 1.2

n = sample of yields = 40

(d) So, test statistics = $\frac{10.2-10}{\frac{1.2}{\sqrt{40} } }$ ~ $t_3_9$

= 1.054

The value of the test statistic is 1.054.

(e) Now, P-value of the test statistics is given by;

P-value = P( $t_3_9$ > 1.054) = 0.1493

Now at 0.05 significance level, the t table gives critical values between -2.0225 and 2.0225 at 39 degree of freedom for two-tailed test. Since our test statistics lies within the critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

(e) Therefore, we conclude that the mean equals the target value of 10 which means that the output can be considered acceptable as it doesn't differs from the target value of 10.

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3 years ago