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4 years ago

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A café offers a lunch special that consists of a sandwich, salad, chips, and a drink. There are 4 sandwich choices, 3 salad choi

ces, 2 chip choices, and 5 drink choices. How many different lunch specials can be ordered?

1 answer:

frutty [35]4 years ago

5 0

Answer:

4

Step-by-step explanation:

there are only for sandwiches.

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On a certain hot summer's day, 391 people used the public swimming pool. The daily prices are $1.25 for children and $2.25 for

igomit [66]

Number of adults = 264

Number of children= 127

Step-by-step explanation:

Let the number of adults be......x

Let the number of children be.....y

The total number of people that used the public swimming pool is given as 391. This means

x+y=391

The total cost per the receipt is : $752.75

This means, 2.25 x+ 1.25 y = 752.75

The two equations are;

x+y=391 ...........(i)

2.25 x+ 1.25 y = 752.75...........(ii)

Making the x variable equal to eliminate x

2.25x + 2.25 y = 879.75

2.25x + 1.25 y = 752.75

Apply subtraction as

1.00 y = 127.00

y=127

Use the value of y in equation (i)

x+y=391

x+127=391

x=391-127=264

Number of adults = 264

Number of children= 127

Learn More

Simultaneous equations: brainly.com/question/12318095

Keywords ; public, swiming pool, prices, daily, adults, children, receipt, admission

#LearnwithBrainly

7 0

4 years ago

An article presents a study of the failure pressures of roof panels. A sample of 15 panels constructed with 8-inch nail spacing

wariber [46]

Answer:

There is sufficient evidence to conclude that six inches spacing has a higher mean failure pressure than 8 inches

Reject $H_0\ at\ 5%$ level of significance

Step-by-step explanation:

From the question we are told that:

Sample size 1 $n_1=15$

Spacing 1 $s_1=8$

Mean 1 $\=x_1=8.48$

Standard deviation 1 $\sigma_1=0.96 kPa$

Sample size 2 $n_2=16$

Spacing 2 $s_2=6$

Mean 1 $\=x_1=9.93$

Standard deviation 1 $\sigma_1=1.02 kPa$

Generally the hypothesis is mathematically given as

Null $H_0:\mu_1-\mu_2$

Alternative $H_a:\mu_1\mu_2$

Generally the equation for pooled estimate is mathematically given by

$S=\sqrt{\frac{(n_1-1)\sigma_1^2+(n_2-1)\sigma_2^2}{n_1+n_2-2} }$

Therefore

$S=\sqrt{\frac{(15-1)0.96^2+(16-1)1.02^2}{15+15-2} }$

$S=0.9905$

Generally the equation for test statistics is mathematically given by

$t=\frac{\=x_1+\=x_2}{S\sqrt{\frac{1}{n _1}+\frac{1}{n_2}}}$

$t=\frac{\=8.48-9.93}{0.9905\sqrt{\frac{1}{15}+\frac{1}{15}}}$

$t=-4.00091$

Therefore From table

$P value=P(t_{15+15-2}=4.00091)$

$P value=0.00041918$

$P$

Therefore

There is sufficient evidence to conclude that six inches spacing has a higher mean failure pressure than 8 inches

Reject $H_0\ at\ 5%$ level of significance

3 0

3 years ago

What is does -27/8+ 2n=n- 5/2 + 5/4m=

enyata [817]

Answer:

m= 8n-7/10

Step-by-step explanation:

n- 5/2 + 5/4 m= -27/8 + 2n

n- 5/2 + 5/4 m-n = -27/8 +2n-4

-5/2 +5 /4m=- 27/8 +n

- 5/2 + 5/4 m + 5/2 = - 27/8+n+ 5/2

5/4 m =n- 7/8

4. 5/4 m=4n -4 . 7/8

5m=4n- 7/2

5m/5 = 4n/5 - 7/2 /5

m= 8n-7/10

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3 years ago

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Bingel [31]

Answer:

i think purex

Step-by-step explanation:

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yawa3891 [41]

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4 years ago

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