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stealth61 [152]
2 years ago
9

Bill has enough money to buy one Half pound of cheeses. How much cheese will he buy?

Mathematics
1 answer:
Zepler [3.9K]2 years ago
4 0

Answer: half pound

Step-by-step explanation: lol

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Help .................
Tema [17]

Answer:

the answer is a

Step-by-step explanation:

5 0
3 years ago
The following quadrilateral is a parallelogram.
Natali5045456 [20]

Answer:

70

Step-by-step explanation:

Angle Q and angle P are supplementary angles, so they add up to 180 degrees.

6x + 4 + 10x = 180

16x + 4 = 180

16x = 176

x = 176/16

x = 11

Put x in angle Q as 11 and solve.

6(11) + 4

66 + 4

= 70

Angle Q measures 70 degrees.

7 0
3 years ago
NEED URGENT HELP<br> ‼️HARD QUESTION ALERT‼️<br> What’s 1+1?
ASHA 777 [7]

Answer:

6

Step-by-step explanation:

5 0
2 years ago
The mean amount of money spent on lunch per week for a sample of 100 students is $21. If the margin of error for the population
Digiron [165]

From the given mean and margin of error, the 99% confidence interval for the mean amount of money spent on lunch per week for all students is:

[$19.5, $22.5].

<h3>How to calculate a confidence interval given the sample mean and the margin of error?</h3>

The confidence interval is given by the sample mean plus/minus the margin of error, hence:

  • The lower bound is the sample mean subtracted by the margin of error.
  • The upper bound is the sample mean added to the margin of error.

For this problem, we have that:

  • The sample mean is of $21.
  • The margin of error is of $1.50.

Hence the bounds are given as follows:

  • Lower bound: 21 - 1.50 = $19.50.
  • Upper bound: 21 + 1.50 = $22.50.

Hence the interval is [$19.50, $22.50].

More can be learned about confidence intervals at brainly.com/question/25890103

#SPJ1

7 0
11 months ago
How to solve this plss frac{(\sqrt{(3)})^{5}}{((\sqrt{(3)})^{-4})}=(\sqrt{(3)})^{(2k+1)​
dusya [7]

Answer:

k = 4

Step-by-step explanation:

Given equation:

\dfrac{(\sqrt{3})^{5}}{(\sqrt{3})^{-4}}=(\sqrt{3})^{(2k+1)}

\textsf{Apply exponent rule} \quad \dfrac{a^b}{a^c}=a^{b-c}:

\implies (\sqrt{3})^{(5-(-4))}=(\sqrt{3})^{(2k+1)}

\implies (\sqrt{3})^{9}=(\sqrt{3})^{(2k+1)}

\textsf{Apply exponent rule} \quad a^{f(x)}=a^{g(x)} \implies f(x)=g(x):

\implies (\sqrt{3})^{9}=(\sqrt{3})^{(2k+1)}

\implies 9=2k+1

\implies 2k=8

\implies k=4

6 0
2 years ago
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