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3 years ago

Journalists are responsible for editing their own articles. A. True B. False

Mathematics

2 answers:

cupoosta [38]3 years ago

7 0

Answer:

true they are responsible for making sure everything is correct

Aleks [24]3 years ago

3 0

Answer:

True

Step-by-step explanation:

At least in my opinion, it’s your journaling is it not? So the journalist should edit it themself.

Hope I helped! :D

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Here's a problem for the minds-good luck!

Art [367]

Answer:

20%

Step-by-step explanation:

Set up a proportion: $\frac{280}{1400} = \frac{x}{100}$
Cross multiply, then divide: 280 × 100 = 28000, 28000 ÷ 1400 = 20
So, 20% were satisfied with their car

I hope this helps!

8 0

3 years ago

Write the contrapositive of the conditional statement. Determine whether the contrapositive is true or false. If it is false, fi

Mila [183]

If you are 16 years old then you are a teenager Question 14

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3 years ago

A soccer game lasts 90 minutes without running into overtime. The coach

KatRina [158]

Answer: 7 Switches will be made.

Step-by-step explanation:

$\frac{90}{12.5}=$ number of switches

$7.2=$ number of switches

however you cannot have a fraction of a switch so you must round it to a whole number, which in this case is 7.

5 0

2 years ago

Does anyone know this answer??

Aleonysh [2.5K]

For this case we have the following equation:

$(2x + 3) ^ 2 + 8 (2x + 3) + 11 = 0$

Let $u = 2x + 3$

We have:

$u ^ 2 + 8u + 11 = 0$

By definition, given an equation of the form $ax ^ 2 + bx + c = 0$

The quadratic formula, to find the solution can be written as:

$x = \frac{-b+/-\sqrt{b ^ 2-4 (a) (c)} }{2(a)}$

In this case we have:

$a = 1\\b = 8\\c = 11$

Substituting in the quadratic formula we have:

See attached image

Answer:

Option B

8 0

3 years ago

Read 2 more answers

(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each

Leviafan [203]

Following are the solution parts for the given question:

For question A:

$\to (n) = 16$

$\to (\bar{X}) = 410$

$\to (\sigma) = 40$

In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

$\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}$

$\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$ When $90\%$ of the confidence interval:

$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

$\to (n) = 500$

$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

$\to C.I= 0.90$

$\to (\alpha) = 0.10$

$\to \frac{\alpha}{2} = 0.05$

Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:

brainly.in/question/16329412

7 0

2 years ago