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9966 [12]
3 years ago
8

The Caesar Cipher of the sentence "THIS MESSAGE IS TOP SECRET."

Mathematics
1 answer:
vagabundo [1.1K]3 years ago
8 0

Answer:  Option 'a' is correct.

Step-by-step explanation:

Since we have given that

"THIS MESSAGE IS TOP SECRET."

We need to use the Caesar cipher.

As in general we always take a shift of 3 keys in Caesar cipher.

so, According to shift of 3 keys it becomes,

T→W

H→K

I→L

S→V

M→P

E→H

A→D

G→J

O→R

C→F

R→U

P→S

So, it can be written as

"THIS MESSAGE IS TOP SECRET."

WKLVP HVVDJ HLVWR SVHFU  HW.

we make a set of 5 letters in a word after Caesar cypher.

Hence, Option 'a' is correct.

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The ice cream vendor 's cold box holds 24 ice cream cups. The
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Answer:

16

Step-by-step explanation:

24/8=3

48/3=16

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3 years ago
Assume that you have a sample of n 1 equals 6​, with the sample mean Upper X overbar 1 equals 50​, and a sample standard deviati
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Answer:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

df=6+5-2=9

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

n_1 =6 represent the sample size for group 1

n_2 =5 represent the sample size for group 2

\bar X_1 =50 represent the sample mean for the group 1

\bar X_2 =38 represent the sample mean for the group 2

s_1=7 represent the sample standard deviation for group 1

s_2=8 represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 > \mu_2

We are assuming that the population variances for each group are the same

\sigma^2_1 =\sigma^2_2 =\sigma^2

The statistic for this case is given by:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

The pooled variance is:

S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

We can find the pooled variance:

S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67

And the pooled deviation is:

S_p=7.46

The statistic is given by:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

The degrees of freedom are given by:

df=6+5-2=9

The p value is given by:

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

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I'm lost.. How do you even start it???
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