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Basile [38]
4 years ago
13

PLEASE HELP!!! PLEASE HELP ASAP!!! I NEED CORRECT ANSWERS ONLY PLEASE!!! I NEED TO FINISH THESE QUESTIONS BEFORE MIDNIGHT TONIGH

T.
Find CD.

Write your answer as an integer or as a decimal rounded to the nearest tenth.

CD =

Mathematics
1 answer:
Nadusha1986 [10]4 years ago
8 0

Answer:

CD = 7.8

Step-by-step explanation:

Given:

Δ DBC is a right angled triangle.

\angle B = 90\°\\\\\angle C = 59\°

BC = 4

We need to find the value of CD

Solution:

Now we know that:

cos\  \theta = \frac{adjacent \ side}{hypotenuse}

So we can say that;

Cos \angle C =\frac{BC}{CD}

Substituting the given values we get;

Cos\ 59 = \frac{4}{CD}\\\\\\CD = \frac{4}{Cos\ 59} = 7.766

Rounding to nearest tenth we get;

CD = 7.8

Hence the Value of CD is 7.8.

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aleta's puppy gained 3/8 pounds each week for 4 weeks. altogether, how much weight did the puppy gained
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1 1/2 pounds is how much her puppy gained altogether
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Miguel has two same size rectangles divivded into the same number of equal parts. One rectangle has 3/4 of the parts shaded, and
ivanzaharov [21]
Bother rectangles could be divided into 8 parts.
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3 years ago
. (0.5 point) We simulate the operations of a call center that opens from 8am to 6pm for 20 days. The daily average call waiting
SashulF [63]

Answer:

The 95% t-confidence interval for the difference in mean is approximately (-2.61, 1.16), therefore, there is not enough statistical evidence to show that there is a change in waiting time, therefore;

The change in the call waiting time is not statistically significant

Step-by-step explanation:

The given call waiting times are;

24.16, 20.17, 14.60, 19.79, 20.02, 14.60, 21.84, 21.45, 16.23, 19.60, 17.64, 16.53, 17.93, 22.81, 18.05, 16.36, 15.16, 19.24, 18.84, 20.77

19.81, 18.39, 24.34, 22.63, 20.20, 23.35, 16.21, 21.73, 17.18, 18.98, 19.35, 18.41, 20.57, 13.00, 17.25, 21.32, 23.29, 22.09, 12.88, 19.27

From the data we have;

The mean waiting time before the downsize, \overline x_1 = 18.7895

The mean waiting time before the downsize, s₁ = 2.705152

The sample size for the before the downsize, n₁ = 20

The mean waiting time after the downsize, \overline x_2 = 19.5125

The mean waiting time after the downsize, s₂ = 3.155945

The sample size for the after the downsize, n₂ = 20

The degrees of freedom, df = n₁ + n₂ - 2 = 20 + 20  - 2 = 38

df = 38

At 95% significance level, using a graphing calculator, we have; t_{\alpha /2} = ±2.026192

The t-confidence interval is given as follows;

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

Therefore;

\left (18.7895- 19.5152 \right )\pm 2.026192 \times \sqrt{\dfrac{2.705152^{2}}{20}+\dfrac{3.155945^2}{20}}

(18.7895 - 19.5125) - 2.026192*(2.705152²/20 + 3.155945²/20)^(0.5)

The 95% CI = -2.6063 < μ₂ - μ₁ < 1.16025996668

By approximation, we have;

The 95% CI = -2.61 < μ₂ - μ₁ < 1.16

Given that the 95% confidence interval ranges from a positive to a negative value, we are 95% sure that the confidence interval includes '0', therefore, there is sufficient evidence that there is no difference between the two means, and the change in call waiting time is not statistically significant.

6 0
3 years ago
Please help! Brainliest:)
Nataliya [291]

Answer:

㏑(\frac{b7}{9})

solution

log10 (\frac{b7}{9 }  = \frac{7log 10 (b)}{9}

steps \frac{log10(b7)}{9}

simplify log 10(b7) : 7log10 (b)

= \frac{7log10(b)}{9}

8 0
2 years ago
Marcus has 4 times the sum of 6 and the number of marbles David has let d represent the number of marbles David has write an alg
Likurg_2 [28]

Answer:

4(6+d)

Step-by-step explanation:

8 0
3 years ago
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