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Vsevolod [243]
3 years ago
8

A cookie recipe calls or 3/4?cups of chocolate chips. You only want to make 2/3 as many cookies as the recipe makes. How many cu

ps of chocolate chips do you need ?
Mathematics
2 answers:
Leni [432]3 years ago
6 0
Less than 1 cup of chocolate chips
Vika [28.1K]3 years ago
3 0
You need half a cup of chocolate chips, that is the answer.
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Tresset [83]
About 14.58% it’s 3499.2 14.59% ends up being around 3501.6
3 0
2 years ago
Carl wants to plant a garden that is 1 1/2 yards long and has an area of 3 1/2 yd.² how wide should the Garden be
lesya692 [45]
L= length= 1 1/2 yards
A= area= 3 1/2 yards^2
w= width

Area= length * width
plug in numbers you know
3 1/2= (1 1/2)(w)
convert to improper fractions
(3*2+1)/2= ((1*2+1)/2)(w)
7/2= 3/2w
divide both sides by 3/2
7/2 ÷ 3/2= w
to divide fractions, multiply by the reciprocal/inverse of 3/2
7/2 * 2/3= w
(7*2)/(2*3)= w
14/6= w
reduce by 2
7/3= width as improper fraction
OR
2 1/3= width as mixed fraction

ANSWER: The width should be 2 1/3 yards wide (or 7/3 yards wide).

Hope this helps! :)
8 0
3 years ago
8x²-x+x²+4x-9x²=18 <br><br>Help meee
malfutka [58]
<span>8x² - x + x² + 4x - 9x² = 18 

3x = 18             | 8x</span>² + x² - 9x² = 0<span>

x = 6                 | divide by 3

Answer: x = 6</span>
5 0
3 years ago
The producer of a certain bottling equipment claims that the variance of all its filled bottles is .027 or less. A sample of 30
o-na [289]

Answer:

p_v = P(\chi^2_{29}>42.963)=1-0.954=0.0459

b. between .025 and .05

Step-by-step explanation:

Previous concepts and notation

The chi-square test is used to check if the standard deviation of a population is equal to a specified value. We can conduct the test "two-sided test or a one-sided test".

\bar X represent the sample mean

n = 30 sample size

s= 0.2 represent the sample deviation

\sigma_o =\sqrt{0.027}=0.164 the value that we want to test

p_v represent the p value for the test

t represent the statistic

\alpha= significance level

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is less than 0.027, so the system of hypothesis are:

H0: \sigma \leq 0.027

H1: \sigma >0.027

In order to check the hypothesis we need to calculate the statistic given by the following formula:

t=(n-1) [\frac{s}{\sigma_o}]^2

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

t=(30-1) [\frac{0.2}{0.164}]^2 =42.963

What is the approximate p-value of the test?

The degrees of freedom are given by:

df=n-1= 30-1=29

For this case since we have a right tailed test the p value is given by:

p_v = P(\chi^2_{29}>42.963)=1-0.954=0.0459

And the best option would be:

b. between .025 and .05

6 0
3 years ago
Find the product of 8(4square rooy7-square root 18
Serggg [28]
what would be the question for this!!
5 0
3 years ago
Read 2 more answers
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