Which problem is even below 4,323 944 291 31

--Marking Brainliest whoever answers first. Please hurry, thank you

Ray Of Light [21]

Answer:

I think it says B; the one in the bottom left. Both B and D are linear functions, but D is a linear function of only y, not x. Therefore, the answer is B.

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Ian brought chocolate And vanilla cupcakes to school for his birthday 75 students decide to take a cupcake and 60% of them fit v

Brrunno [24]

Answer:

45 kids picked vanilla cupcake

Step-by-step explanation:

We know that there are 75 students in total who took a cupcake. If 60% took vanilla then this is a simple equation.

75 * 60% or .60 = 45.

45 kids picked vanilla cupcake

6 0

3 years ago

Evaluate of function h(3a)=x^4+2a^2+4

amid [387]

If you would like to evaluate function h(x) = x^4 + 2 * a^2 + 4 when x = 3a, you can do this using the following steps:

x = 3a
h(x) = x^4 + 2 * a^2 + 4
h(3a) = (3a)^4 + 2 * a^2 + 4 = 3^4 * a^4 + 2 * a^2 + 4 = 81 * a^4 + 2 * a^2 + 4

The correct result would be h(3a) = 81 * a^4 + 2 * a^2 + 4.

6 0

3 years ago

A simple random sample from a population with a normal distribution of 98 body temperatures has x =98.90 °F and s =0.68°F. Const

Ludmilka [50]

Answer:

$0.609 \leq \sigma \leq 0.772$

And the best conclusion would be:

D. This conclusion is safe because 1.40 °F is outside the confidence interval.

Step-by-step explanation:

1) Data given and notation

s=0.68 represent the sample standard deviation

$\bar x =98.90$ represent the sample mean

n=98 the sample size

Confidence=90% or 0.90

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=98-1=97$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical values.