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3 years ago

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Lilly wants to get some granola bars. At Kroger you can get 6 strawberry and 4 cherry together for $6.90. At Sams club you can g

et 5 strawberry and 9 cherry for $9.15. How much is each granola bar individually?

1 answer:

Charra [1.4K]3 years ago

5 0

each granola bar at kroger is 69¢, each granola bar at sams is 65¢

Step-by-step explanation:

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Julie and Tim are both cable subscribers. Julie pays $70 per month plus an additional $4.25 for each movie she orders. Tim pays

Kaylis [27]

Answer:

no of movies (m), by Tim = 13

Step-by-step explanation:

Julie payment = constant 70 + 4.25 per additional movie = 70 + 4.25 m

Tim payment = constant 49.5 + 5.5 per additional movie = 49.5 + 5.5 m

Julie order = 12 movies. So, final payment = 70 + 4.25 (12) = 70 + 51 = 121

Equalising Tim's equation, for finding Tim's movies for same final amount , 121 = 49.5 + 5.5 m → 5.5m = 121 - 49.5 = 71.5 → m = 71.5 / 5.5 = 13

4 0

3 years ago

Natalia paid $38.95 for three medium-sized pizzas and a salad. Solve an equation to find the cost of one pizza,p, if Natalia pai

Vedmedyk [2.9K]

One Pizza is 8.99 hope this helps

7 0

3 years ago

Read 2 more answers

PLS ANSWER AS QUICKLY AS POSSIBLE PLSSSSSSSS

igor_vitrenko [27]

Answer:120

DCA=25+35=60

X=DCA+60=60+60=120

Step-by-step explanation:

5 0

2 years ago

A spinner is divided into eight equal-sized sections, numbered from 1 to 8, inclusive.

ss7ja [257]

A.
The spinner can land on 1, 2, 3, 4, 5, 6, 7, or 8, so the sample space is {1, 2, 3, 4, 5, 6, 7, 8}.
Choice A applies.

B.
1, 2, 3 are part of the sample space, so choice B applies.

C.
9 is not in the sample space, so choice C does not apply.

D.
4 is in set A. 4 is not less than 4, so choice D does not apply.

E.
The set of spinning an odd number is {1, 3, 5, 7}.
Removing these elements from the original sample space leaves the complement (2, 4, 6, 8}. Choice E applies.

Answer: A, B, E

7 0

3 years ago

Read 2 more answers

A machine produces 3-inch nails. A sample of 10 nails is obtained and the lengths determined. After some calculation, the sample

gtnhenbr [62]

Answer:

<em>The 90% confidence interval for the population mean length of nails produced by the machine.</em>

<em>(2.91562, 3.04438)</em>

Step-by-step explanation:

<u><em>Step:1</em></u>

Given that the size of the sample 'n' = 10

Given that the mean of the sample (x⁻) = 2.98

Given that the standard deviation of the sample (s) = 0.09

Degrees of freedom = n-1 = 10 -1 = 9

t₀.₁₀ = 2.2621

<u><em>Step:2</em></u>

<em>The 90% confidence interval for the population mean length of nails produced by the machine.</em>

$(x^{-} - t_{9 , 0.10} \frac{S}{\sqrt{n} } , x^{-} + t_{9,0.10} \frac{S}{\sqrt{n} } )$

$(2.98 - 2.2621(\frac{0.09}{\sqrt{10} } , 2.98 + 2.262 \frac{0.09}{\sqrt{10} } )$

(2.98 - 0.06438 , 2.98 + 0.06438)

(2.91562 , 3.04438)

<u><em>Final answer:-</em></u>

<em>The 90% confidence interval for the population mean length of nails produced by the machine.</em>

<em>(2.91562, 3.04438)</em>

4 0

3 years ago