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3 years ago

Please help me thank you I appreciate it

Mathematics

2 answers:

Westkost [7]3 years ago

7 0

90-42= 48

It’s a right angle.
Right angles are 90 degrees

LuckyWell [14K]3 years ago

4 0

Answer:

angle FDE+angle CDF=90° (given CDE=90°)

angle CDF=90-42

angle CDF=48°

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Ratling [72]

Answer:

let x as a cointerior angle and find the

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Of the dogs at the dog show,

Olin [163]

So, let's say that all the dogs had a number of x.

then 1/12 x were the Hound Group dogs.

then, out of the Hound Group,, that is 1/12 x, 3/10 were English Foxhounds.

For this, we need to multiply the fractions:

$\frac{1}{12}x* \frac{3}{10} = \frac{3}{120}x =\frac{1}{40}x$

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3 years ago

Read 2 more answers

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EastWind [94]

Answer:

the answer is 30 with 75 kids voting for drinks in total

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3 years ago

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Help please!!!!!!!!!!!!!!

marshall27 [118]

Step-by-step explanation:

as in the other problem, the table describes mutually exclusive attributes.

so, we can simply sum the numbers up to get the total number of students :

3+9+2+16 = 30 students.

so, when picking a random student, we have 30 possibilities.

now, how many of those 30 have the desired attributes (to have a brother AND a sister) ?

I see 3 in the table.

so, the probability to pick a student that has a brother and a sister is

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3 0

2 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

3 years ago