All categories

3 years ago

Click on image and answer with explanation.

Mathematics

1 answer:

marshall27 [118]3 years ago

3 0

60% of 90 is 54 by doing 90*60/100
2/3 of 150 is 100 because 150 divided by 3 times 2 is 100
and dawn sold 46 cookies.

if you add all the cookies sold up, you get 200 cookies.
Dawn having sold 46 cookies out of the 200, if you want to have a percentage, you have to change that 200 which is the total into 100.

So 200/2=100 and 46/2=23

Dawn sold 23 cookies for every 100 cookies sold so 23%

You might be interested in

Choose the equation of the horizontal line that passes through the part point (1,-5)

vladimir2022 [97]

We know is a horizontal line, so, if it passes through 1,-5, it also passes through "whatever", -5, like 20, -5 or 1000000, -5, or -100000000, -5 and so on.

so, let's pick another point say -7, -5, check the picture below, and let's check about the equation that runs through it,

$\bf \begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~ 1 &,& -5~) 
% (c,d)
&&(~ -7 &,& -5~)
\end{array}
\\\\\\
% slope = m
slope = m\implies 
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-5-(-5)}{-7-1}
\\\\\\
\cfrac{-5+5}{-7-1}\implies \cfrac{0}{-8}\implies 0
\\\\\\
% point-slope intercept
\stackrel{\textit{point-slope form}}{y- y_1= m(x- x_1)}\implies y-(-5)=0(x-1)
\\\\\\
y+5=0\implies y=-5$

4 0

3 years ago

Read 2 more answers

From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean leng

alexandr402 [8]

Complete Question

From a random sample of size 18, a researcher states that (11.1, 15.7) inches is a 90% confidence interval for mu, the mean length of bass caught in a small lake. A normal distribution was assumed. Using the 90% confidence interval obtain:

a. A point estimate of $\mu$ and its 90% margin of error.

b. A 95% confidence interval for $\mu$ .

Answer:

$\= x = 13.4$ . $E = 2.3$

$10.7 < \mu < 16.1$

Step-by-step explanation:

From the question we are told that

The sample size is n = 18

The 90% confidence interval is (11.1, 15.7)

Generally the point estimate of $\mu$ is mathematically evaluated as

$\= x = \frac{11.1 + 15.7 }{2}$

=> $\= x = 13.4$

Generally the margin of error is mathematically evaluated as

$E = \frac{15.7 - 11.1}{2 }$

=> $E = 2.3$

From the question we are told the confidence level is 90% , hence the level of significance is

$\alpha = (100 - 90 ) \%$

=> $\alpha = 0.10$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the equation for the lower limit of the confidence interval is

$\= x - Z_{\frac{\alpha }{2} } * \frac{s}{\sqrt{18} } = 11.1$

=> $13.4 - 0.3877 s = 11.1$

=> $s = 5.932$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E = 1.96 * \frac{5.932}{\sqrt{18} }$

=> $E = 2.7$

Generally 95% confidence interval is mathematically represented as

$\= x -E < \mu < \=x +E$

=> $13.4 - 2.7 < \mu < 13.4 + 2.7$

=> $10.7 < \mu < 16.1$

3 0

3 years ago

In a certain year, when she was a high school senior, Idonna scored 671 on the mathematics part of the SAT. The distribution of

goldfiish [28.3K]

Answer:

Idonna's standardized score is 1.41.

Jonathan's standardized score is 0.55.

A.) Idonna's score is higher than Jonathan's

Step-by-step explanation:

Z-score:

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Idonna scored 671 on the mathematics part of the SAT. The distribution of SAT math scores in that year was Normal with mean 509 and standard deviation 115.

This means that her standardized score is Z when $X = 671, \mu = 509, \sigma = 115$ . So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{671 - 509}{115}$

$Z = 1.41$

Idonna's standardized score is 1.41.

Jonathan took the ACT and scored 24 on the mathematics portion. ACT math scores for the same year were Normally distributed with mean 21.1 and standard deviation 5.3 .

This means that his standardized score is Z when $X = 24, \mu = 21.1, \sigma = 5.3$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{24 - 21.1}{5.3}$

$Z = 0.55$

Jonathan's standardized score is 0.55.

Due to the higher z-score, Iddona's has a higher score.

5 0

2 years ago

Mark save $258 more than Neil. In fact Mark saved $7 for every $4 Neil saved. How much money did Mark say?

olasank [31]

Let

x---------> amount of money saved by Mark

x---------> amount of money saved by Neil

we know that

x=y+258-------> equation 1

x/y=(7/4)-------> equation 2

substitute equation 1 in equation 2

(y+258) /y=7/4--------> 4*(y+258)=7*y------> 4y+1032=7y

7y-4y=1032--------> 3y=1032------> y=$344

x=344+258------>x=$602

the answer is

$602

6 0

3 years ago

Use rulers and compasses to draw a line which is perpendicular to line AB at point C

netineya [11]

Answer:

Draw a line AB

Take a point C on AB

Fix the tip of compass at C and with a fix radius, cut AB on both sides of C

Mark those points D and E

Fix the tip of compass on D, and with the same radius, draw an arc above C

Do the same with E

Those two arcs will intersect at a point, mark the point F

Make a line passing through both C and F

Line CF is perpendicular to AB

8 0

2 years ago