If the relative frequency of getting a blue on a spinner is 0.1 how many reds would you expect to get in 50 spins?

Which method listed below could not be used to prove that two triangles are congruent? a. Prove all three sets of corresponding

Maurinko [17]

Answer:

All corresponding sides and angles will be congruent

Step-by-step explanation:

7 0

3 years ago

Anna surveyed 300 of the students in her school about their favorite color. 186 students said their favorite color was red. What

MArishka [77]

Answer:

62%

Step-by-step explanation:

I took a fraction and made a decimal which was made into a percentage.

186 ÷ 300 = .62

Check: 62% • 300 = 186

6 0

3 years ago

<img src="https://tex.z-dn.net/?f=%5Cfrac%7B3%2F7%7D%7B7%5Csqrt%7B10%7D%2F2%20%7D" id="TexFormula1" title="\frac{3/7}{7\sqrt{10}

VashaNatasha [74]

Multiply the numerator and denominator by 7×2 = 14 to eliminate the denominators of those fractions:

$\dfrac{\dfrac37}{\dfrac{7\sqrt{10}}2}\times\dfrac{14}{14}=\dfrac{3\times2}{7\sqrt{10}\times7}=\dfrac6{49\sqrt{10}}$

Rationalize the denominator by multiplying both numerator and denominator by √10:

$\dfrac6{49\sqrt{10}}\times\dfrac{\sqrt{10}}{\sqrt{10}}=\dfrac{6\sqrt{10}}{49(\sqrt{10})^2}=\dfrac{6\sqrt{10}}{49\times10}=\dfrac{6\sqrt{10}}{490}$

Lastly, cancel the common factor of 2 in both the numerator and denominator (which comes from 6 = 2×3 and 490 = 2×245):

$\dfrac{6\sqrt{10}}{490}=\dfrac{3\sqrt{10}}{245}}$

8 0

3 years ago

Could someone help me please?

fenix001 [56]

Answer:

a. 7.2

b. 6

Step-by-step explanation:

Since the side lengths of similar polygons are proportional to each other, therefore:

$\frac{b}{10} = \frac{9}{15} = \frac{a}{12}$

Thus, let's find b using:

$\frac{b}{10} = \frac{9}{15}$

Cross multiply

$b*15 = 10*9$

$15b = 90$

Divide both sides by 15

b = 6

Let's find a using:

$\frac{9}{15} = \frac{a}{12}$

Cross multiply

9*12 = 15*a

108 = 15a

Divide both sides by 15

a = 7.2

5 0

3 years ago

A researcher wishes to estimate the average blood alcohol concentration (BAC) for drivers involved in fatal accidents who are f

Mnenie [13.5K]

Answer:

A 90% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC is [0.143, 0.177] .

Step-by-step explanation:

We are given that a researcher randomly selects records from 60 such drivers in 2009 and determines the sample mean BAC to be 0.16 g/dL with a standard deviation of 0.080 g/dL.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

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

where, $\bar X$ = sample mean BAC = 0.16 g/dL

s = sample standard deviation = 0.080 g/dL

n = sample of drivers = 60

$\mu$ = population mean BAC in fatal crashes

<em>Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation. </em>

So, a 90% confidence interval for the population mean, $\mu$ is;

P(-1.672 < $t_5_9$ < 1.672) = 0.90 {As the critical value of t at 59 degrees of

freedom are -1.672 & 1.672 with P = 5%} P(-1.672 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.672) = 0.90

P( $-1.672 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.672 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.672 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.672 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

<u>90% confidence interval for</u> $\mu$ = [ $\bar X-1.672 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.672 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $0.16-1.672 \times {\frac{0.08}{\sqrt{60} } }$ , $0.16+1.672 \times {\frac{0.08}{\sqrt{60} } }$ ]

= [0.143, 0.177]

Therefore, a 90% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC is [0.143, 0.177] .

7 0

4 years ago