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nikklg [1K]
3 years ago
15

Ragon, who lives on this planet, does a survey and finds that her colony of 237 contains 45 black-footed, one-headed Mizjigs; 69

red-footed, two-headed Mizjigs; and 115 one-headed Mizjigs.
How many black footed Mizjigs are there in Ragon's colony?

Mathematics
1 answer:
Gemiola [76]3 years ago
4 0

The answer is <u>A. 98</u>.

Because when you subtract 115 - 45 it equals 70 then you add 70 + 69 which equals 139 then you subtract 237 - 139 it equals <u>98</u> which is your answer!

<h2>      Hope this helps!</h2>
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Which statement is true regarding the graphed functions?
Paraphin [41]

Answer:

f(0)=g(0)

Step-by-step explanation:

The problem is asking for where the output (y value) of one function, f(x), matches the output of another function, g(x), otherwise called an intersection. You can see that the two lines intersect at 0 on the x axis in the graph, therefore f(0)=g(0) meaning the outputs of f(x) are the same as g(x) at x value 0.

5 0
2 years ago
A simple random sample of 20 days in which Parsnip ate seeds was selected, and the mean amount of time it took him to eat the se
bearhunter [10]

Answer:

(14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63

(14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83

And the confidence interval for this case -1.63 \leq \mu_1 -\mu_2 \leq 2.83

Step-by-step explanation:

We know the following info from the problem

\bar X_1 = 14.5 sample mean for the group 1

s_1 = 3.98 the standard deviation for the group 1

n_1= 20 the sample size for group 1

\bar X_2 = 13.9 sample mean for the group 2

s_1 = 4.03 the standard deviation for the group 2

n_2= 17 the sample size for group 2

We have all the conditions satisifed since we have random samples.

We want to construct a confidence interval for the true difference of means and the correct formula for this case is:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}

The degrees of freedom are given :

df = n_1 +n_2- 2 = 20+17-2=35

The confidence level is 0.9 or 90% and the significance level is \alpha=1-0.9=0.1 and \alpha/2 = 0.05 and the critical value for this case is:

t_{\alpha/2} = 1.69

And replacing the info given we got:

(14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63

(14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83

And the confidence interval for this case -1.63 \leq \mu_1 -\mu_2 \leq 2.83

5 0
3 years ago
Sara is cutting bows out of ribbon which
Andrei [34K]
She can make at least 1 because if u multiple 1/4 and 5 it will give u a decimal as 1.25 or 5/4, or 1
3 0
3 years ago
Read 2 more answers
How did my teacher get 11 for #17
wel
Complementary angles ⇒ the angles add up to 90°

∠A and ∠B are complementary ⇒ ∠A + ∠B = 90°

∠A + ∠B = 90°

Plug in the values for ∠A and ∠B:
(3x - 8) + (5x + 10)  = 90

Open the brackets:
3x - 8 + 5x + 10 = 90

Combine like terms:
8x + 2 = 90

Subtract 2 from both sides:
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Divide both sides by 8:
x = 11

Answer: x = 11
5 0
3 years ago
Read 2 more answers
Can someone help me with this :( .
andrew11 [14]

Answer:

Which question?

Step-by-step explanation:

8 0
3 years ago
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