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zhuklara [117]
2 years ago
10

A student purchases 6 different colored folders for her classes (red, blue, yellow, green, purple, and black). She randomly choo

ses two folders to use for her first two classes. What is the probability that her first choice is red and her second choice is blue?
Mathematics
2 answers:
kondor19780726 [428]2 years ago
5 0
Red is 1 out of 6 as there are 6 colors
blue is 1 out of 5 as there are 5 remaining colors to pick from
sveticcg [70]2 years ago
5 0

Answer:

The answer is

Step-by-step explanation:

1/30 or 1 out of 30

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The rabbit population on a small island is observed to be given by the function P(t) = 130t − 0.3t^4 + 1100 where t is the time
Montano1993 [528]
The maximum occurs when the derivative of the function is equal to zero.
P(t)=-0.3t^{4}+130t+1100 \\ P'(t)=-1.2t^{3}+130 \\ 0=-1.2t^{3}+130 \\ 1.2t^{3}=130 \\ t^{3}= \frac{325}{3}  \\ t=4.76702
Then evaluate the function for that time to find the maximum population.
P(t)=-0.3t^{4}+130t \\ P(4.76702)=-0.3*4.76702^{4}+130*4.76702+1100 \\ P(4.76702)=1564.79201
Depending on the teacher, the "correct" answer will either be the exact decimal answer or the greatest integer of that value since you cannot have part of a rabbit.
7 0
2 years ago
Complete the similarity statement for the two triangles shown. Enter your answer in the box. △XBR∼△ Two similar triangles B R X
Andreyy89

Answer:  

Here, BRX and NJY are two triangles in which,

BR = 30 cm, RX = 40 cm, BX = 60 cm, NJ = 15 cm, JY = 20 cm and NY = 30 cm,

Also, m∠B = m∠N, m∠R = m∠J and m∠X = m∠Y,

By the property of congruence,

\angle B \cong \angle N, \angle R \cong \angle J and \angle X \cong \angle Y

Thus, By AAA similarity postulate,

\triangle BRX\sim \triangle NJY

Hence, proved.

5 0
2 years ago
what are the solution(s) to the quadratic equation x2 – 25 = 0? x = 5 and x = –5 x = 25 and x = –25 x = 125 and x = –125 no real
sattari [20]
The left hand side expression of the given equation is a difference of two squares. The first term, x², is a square of x and the second term, 25 is the square of 5. The factors of the expression are (x - 5) and (x + 5).
                                   (x - 5)(x + 5) = 0
The values of x from the equation above are x = -5 and x = 5.
3 0
2 years ago
Read 2 more answers
What is greater than 1.45? A. 0.009 B. 0.019 C. 0.0032 D. 0.0177
8_murik_8 [283]

Answer:

none of them are greater than 1.45

8 0
2 years ago
Read 2 more answers
Marketing companies have collected data implying that teenage girls use more ring tones on their cell phones than teenage boys d
ASHA 777 [7]

Answer:

The p-value here is 0.0061, which is very small and we have evidence that the girls' mean is higher than the boys' mean.

Step-by-step explanation:

We suppose that the two samples are independent and normally distributed with equal variances. Let \mu_{1} be the mean number of ring tones for girls, and \mu_{2} the mean number of ring tones for boys.

We want to test H_{0}: \mu_{1}-\mu_{2} = 0 vs H_{1}: \mu_{1}-\mu_{2} > 0 (upper-tail alternative).

The test statistic is

T = \frac{\bar{X}_{1}-\bar{X}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}

where

S_{p} = \sqrt{\frac{(n_{1}-1)S_{1}^{2}+(n_{2}-1)S_{2}^{2}}{n_{1}+n_{2}-2}}.

For this case, n_{1}=n_{2}=20, \bar{x}_{1}=3.2, s_{1}=1.5, \bar{x}_{2}=2.2, s_{2}=0.8.

s_{p} = \sqrt{\frac{(19)(1.5)^{2}+(19)(0.8)^{2}}{38}} = 1.2021 and the observed value is

t = \frac{3.2-2.2}{1.2021\sqrt{1/20+1/20}} = \frac{1}{0.3801} = 2.6309.

We can compute the p-value as P(T > 2.6309) where T has a t distribution with 20 + 20 - 2 = 38 degrees of freedom, so, the p-value is 0.0061. Because the p-value is very small, we can reject the null hypothesis for instance, at the significance level of 0.05.

3 0
2 years ago
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