Please help will mark you brainist

Pastries made out of filo dough are brushed with either olive oil or butter (but not both). Pastries made out of shortcrust doug

givi [52]

Answer:

3X/20 (option a) of the pastries submitted by Rashid and Mikhail were brushed with butter

Step-by-step explanation:

Rashid pastries (R)

Mikhail pastries (M)

Rashid and Mikhail submitted a total of x pastries

R+M=x (I)

Rashid made 2/3 as many pastries as Mikhail

(2/3)*R=M (II)

Using II in I

R+(2/3)*R=x

(5/3)*R = x

R=(3/5)*x (III)

Using III in I

(3/5)*x+M=x

M=x-(3/5)*x

M=(2/5)*x (IV)

Mikhail filo dough (MF)

Mikhail shortcrust dough (MS)

Rashid filo dough (RF)

Rashid shortcrust dough (RS)

Mikhail used filo dough for all of his pastries

MF=M

MS=0

Rashid used shortcrust dough for all of his pastries

RS=R

RF=0

Filo dough (FD)

FD=RF+MF=0+MF=MF=M (V)

5/8 of the filo dough pastries were brushed with olive oil

pastries brushed with olive oil (OI)

(5/8)*FD=OI

Using V

(5/8)*M=OI

Using IV

(5/8)*(2/5)*x=OI

(1/4)*x=OI (VI)

pastries brushed with butter (B)

Pastries made out of filo dough are brushed with either olive oil or butter (but not both)

FD=OI+B

B=FD-OI

Using V and VI

B= M - (1/4)*x

Using IV

B = (2/5)*x - (1/4)*x

B= (3/20)*x

3X/20 (option a) of the pastries submitted by Rashid and Mikhail were brushed with butter

5 0

3 years ago

Please help me out!!

solniwko [45]

The slope is 1 because it is parallel and y=x+b, so, 2=-3+b and b=5
y=x+5

8 0

3 years ago

23. A certain type of gasoline is supposed to have a mean octane rating of at least 90. Suppose measurements are taken on of 5 r

Nostrana [21]

Answer:

a) $p_v =P(Z$

b) Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

c) $P(\bar X >90)=1-P(Z$

$1-P(Z$

d) For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

Step-by-step explanation:

Previous concepts and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

We can calculate the sample mean and deviation with the following formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}$

$\bar X=88.96$ represent the sample mean

$s=1.011$ represent the sample standard deviation

n=5 represent the sample selected

$\alpha$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 90, the system of hypothesis would be:

Null hypothesis: $\mu \geq 90$

Alternative hypothesis: $\mu < 90$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{88.96-90}{\frac{0.8}{\sqrt{5}}}=-2.907$

Part a

P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=5-1 = 4$

Then since is a left tailed test the p value would be:

$p_v =P(Z$

Part b

Since $p_v$ we can reject the null hypothesis at the significance level given. So based on this we can commit type of Error I.

Because Type I error " is the rejection of a true null hypothesis".

Part c

We want this probability:

$P(\bar X$

The best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

If we apply this formula to our probability we got this:

$P(\bar X >90)=1-P(Z$

$1-P(Z$

Part d

For this case we need a z score that accumulates 0.02 of the area on the right tail and 0.98 on the left tail and this value is z=2.054

And we can use this formula:

$2.054=\frac{90-89}{\frac{0.8}{\sqrt{n}}}$

And if we solve for n we got:

$n = (\frac{2.054 *0.8}{1})^2=2.7 \approx 3$

7 0

3 years ago

Some students reported how many pets they had. The dot plot shows the data collected: A dot plot is titled Students Pets and lab

hichkok12 [17]

Answer:B

Step-by-step explanation:I just know

4 0

3 years ago

In a computer game, a bird flies at a height of 1,281 feet. At the same time, a fish swims 282 feet below sea level. What is the

Nana76 [90]

Answer:

The difference is 1,563 feet.

Step-by-step explanation:

1,281+282=1,563

5 0

3 years ago