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3 years ago

12

Angie baked 100 cookies, and 20 brownies. She wants to split them into equal groups for the bake sale .Each group must have the

same number of cookies and brownies , with none leftover. What is the greatest number of groups she can make?

1 answer:

densk [106]3 years ago

6 0

20 is the greatest common factor Each group of 20 will have 1 brownie & 5 cookies

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Show work please please

Bess [88]

Answer:

can you give a word problem and not just a picture

Step-by-step explanation:

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3 years ago

The company is building a scale model of the theater's main show tank for an investor's presentation. Each dimension will be mad

Ede4ka [16]

Answer:

This means the mock-up tank has a volume 216 times smaller than the initial tank.

Step-by-step explanation:

The ratio between volume is given by

$\frac{V_2}{V_1}=\frac{(1/6)L^{3} }{L^{3}} =\frac{1}{216}$

This means the mock-up tank has a volume 216 times smaller than the initial tank.

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3 years ago

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Brendan has 33 coins in his pocket, all of which are dimes and quarters. if the total value of his change is 615 cents, how many

Amanda [17]

If all were dimes, the value would be $3.30. Brendan has $2.85 more than that. Each quarter that replaces a dime increases the value by $0.15, so Brendan must have $2.85/$0.15 = 19 quarters.

Brendan has 14 dimes and 19 quarters.

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2 years ago

Meryl spend a total of 68.82 for 2 pairs of sneakers with the same cost . The sale tax is 5.32 . Meryl also use a coupon for 3.0

Nimfa-mama [501]

$68.82
$68.82 - $5.32
$63.50 - $3.00
$60.50 / 2
$30.25 each pair of shoes

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3 years ago

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The calibration of a scale is to be checked by weighing a 13 kg test specimen 25 times. Suppose that the results of different we

Natali5045456 [20]

Solution :

a).

Given : Number of times, n = 25

Sigma, σ = 0.200 kg

Weight, μ = 13 kg

Therefore the hypothesis should be tested are :

$H_0 : \mu = 13$

$H_a : \mu \neq 13$

b). When the value of $\overline x = 12.84$

Test statics :

$Z=\frac{(\overline x - \mu)}{\frac{\sigma}{\sqrt n}}$

$Z=\frac{(14.82-13)}{\frac{0.2}{\sqrt {25}}}$

$=\frac{1.82}{0.04}$

= 45.5

P-value = 2 x P(Z > 45.5)

= 2 x 1 -P (Z < 45.5) = 0

Reject the null hypothesis if P value < α = 0.01 level of significance.

So reject the null hypothesis.

Therefore, we conclude that the true mean measured weight differs from 13 kg.

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3 years ago