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3 years ago

Is number 11 help plz

Mathematics

1 answer:

LenKa [72]3 years ago

3 0

A) 1 gallon/5 pints
b) 3 quarts/5 pints
c) 5 pints/8 ounces
d)3 quarts/1 gallon
e) 8 ounce/1 gallon

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Find the length of side y

Serggg [28]

Answer:

$\large\boxed{y=76.3359\ cm\approx76.3\ cm}$

Step-by-step explanation:

Use cosine:

$cosine=\dfrac{adjacent}{hypotenuse}$

We have

$adjacent=73\ cm\\hypotenuse=y$

and the angle 17°.

$\cos17^o\approx0.9563$

Substitute:

$\dfrac{73}{y}=0.9563$ convert the decimal to the fraction

$\dfrac{73}{y}=\dfrac{9563}{10000}$ cross multiply

$9563y=730000$ divide both sides by 9563

$y=\dfrac{730000}{9563}$

$y\approx76.3359$

8 0

3 years ago

There are 45 students who play a woodwind instrument in the school band. Of these, 18 play the saxophone. What percent of these

Mashutka [201]

Answer:

40% students play the saxophone.

Step-by-step explanation:

Given:

Total Number of students who play woodwind instrument = 45

Number of students who play saxophone = 18

We need to find the percent of students who play saxophone.

Solution:

Now we can say that;

To find the percent of students who play saxophone we will divide Number of students who play saxophone by Total Number of students who play woodwind instrument and the multiply by 100.

framing in equation form we get;

percent of students who play saxophone = $\frac{18}{45}\times100= 40\%$

Hence 40% students play the saxophone.

3 0

3 years ago

A survey was conducted that asked 1003 people how many books they had read in the past year. Results indicated that x overbar eq

Sergio [31]

Answer:

The 99% confidence interval would be given (11.448;14.152).

Step-by-step explanation:

1) Important concepts and notation

A confidence interval for a mean "gives us a range of plausible values for the population mean. If a confidence interval does not include a particular value, we can say that it is not likely that the particular value is the true population mean"

$s=16.6$ represent the sample deviation

$\bar X=12.8$ represent the sample mean

n =1003 is the sample size selected

Confidence =99% or 0.99

$\alpha=1-0.99=0.01$ represent the significance level.

2) Solution to the problem

The confidence interval for the mean would be given by this formula

$\bar X \pm z_{\alpha/2} \frac{s}{\sqrt{n}}$

We can use a z quantile instead of t since the sample size is large enough.

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.